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Physics 5 Marks Questions

Newton’s Laws of Motion · MP Board (MPBSE) STD 12 Science (Madhya Pradesh - English Medium). 45 questions.

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Question 15 Marks
Suppose the ceiling in the previous problem is that of an elevator which is going up with an acceleration of $2.0m/s^2.$ Find the elongations.
Answer
$a = 2m/s^2 kl - (2g + 2a) = 0$
$\Rightarrow kl = 2g + 2a = 2 \times 9.8 + 2 \times 2 = 19.6 + 4 = 23.6$
$\Rightarrow\text{I}=\frac{23.6}{100}=0.236\text{m}=0.24\text{m}$

When 1kg body is added total mass $(2 + 1)kg = 3kg.$
elongation be $l_1\   kl_1 = 3g + 3a = 3 \times 9.8 + 6 = 33.4$
$\Rightarrow\text{I}_1=\frac{33.4}{100}=0.0334=0.36$
Further elongation $= l_1 - l = 0.36 - 0.24 = 0.12m.$ 
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Question 25 Marks
An empty plastic box of mass m is found to accelerate up at the rate of $\frac{\text{g}}{6}$ when placed deep inside water. How much sand should be put inside the box so that it may accelerate down at the rate of $\frac{\text{g}}{6}?$
Answer
When the box is accelerating upward,
$\text{U}-\text{mg}-\text{m}\Big(\frac{\text{g}}{6}\Big)=0$
$\Rightarrow\text{U}=\text{mg}+\frac{\text{mg}}{6}=\text{m}\Big\{\text{g}+\Big(\frac{\text{g}}{6}\Big)\Big\}=7\frac{\text{mg}}{7} \ ...(\text{i})$
$\Rightarrow\text{m}=\frac{6\text{U}}{7\text{g}}.$
When it is accelerating downward, let the required mass be M.$\text{U}-\text{Mg}+\frac{\text{Mg}}{6}=0$
$\Rightarrow\text{U}=\frac{6\text{Mg}-\text{Mg}}{6}=\frac{5\text{Mg}}{6}\Rightarrow\text{M}=\frac{6\text{U}}{5\text{g}}$
Mass to be added $=\text{M}-\text{m}=\frac{6\text{U}}{5\text{g}}-\frac{6\text{U}}{7\text{g}}=\frac{6\text{U}}{\text{g}}\Big(\frac{1}{5}-\frac{1}{7}\Big)$
$=\frac{6\text{U}}{\text{g}}\Big(\frac{2}{35}\Big)=\frac{12}{35}\Big(\frac{\text{U}}{\text{g}}\Big)$
$=\frac{12}{35}\Big(\frac{7\text{mg}}{6}\times\frac{1}{\text{g}}\Big)$ from (i)
$=\frac{2}{5}\text{m}.$
$\therefore$ The mass to be added is $\frac{2\text{m}}{5}.$
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Question 35 Marks
The elevator shown in figure is descending with an acceleration of $2m/s^2$. The mass of the block A is 0.5kg. What force is exerted by the block A on the block B?
Answer
From the free body diagram
$\therefore R + 0.5 × 2 - w = 0$
$⇒ R = w - 0.5 × 2 $
$⇒ 0.5 (10 - 2) = 4N.$
So, the force exerted by the block A on the block B, is 4N.
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Question 45 Marks
The monkey B shown in figure is holding on to the tail of the monkey A which is climbing up a rope. The masses of the monkeys A and B are 5kg and 2kg respectively. If A can tolerate a tension of 30N in its tail, what force should it apply on the rope in order to carry the monkey B with it? Take $g = 10m/s^2$
Answer
Suppose A move upward with acceleration a, such that in the tail of A maximum tension 30N produced.

$T - 5g - 30 - 5a = 0 ...(i)$
$\Rightarrow T = 50 + 30 + (5 \times 5) = 105N$ (max)
$30 - 2g - 2a = 0 ...(ii)$
$\Rightarrow 30 - 20 - 2a = 0$
$\Rightarrow a = 5m/s^2​​​​​​​$
So, A can apply a maximum force of 105N in the rope to carry the monkey B with it.
For minimum force there is no acceleration of monkey ‘A’ and B.
$\Rightarrow a = 0$ Now equation (ii) is
$T'_1 - 2g = 0 \Rightarrow T'_1 = 20N$
​​​​​​​(wt. of monkey B) Equation (i) is
$T - 5g - 20 = 0 [$As $T'_1 = 20N]$
$​​​​​​​\Rightarrow T = 5g + 20 = 50 + 20 = 70N.$
$\therefore$ The monkey A should apply force between 70N and 105N to carry the monkey B with it.
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Question 55 Marks
The force of buoyancy exerted by the atmosphere on a balloon is B in the upward direction and remains constant. The force of air resistance on the balloon acts opposite to the direction of velocity and is proportional to it. The balloon carries a mass M and is found to fall down near the earth's surface with a constant velocity v. How much mass should be removed from the balloon so that it may rise with a constant velocity v?
Answer
Let, the air resistance force is F and Buoyant force is B. Given that$\text{F}_{\text{a}}\propto\text{v},$ v where → velocity
⇒ Fa = kv, where k → proportionality constant.
When the balloon is moving downward, B + kv = mg …(i)$\Rightarrow\text{M}=\frac{\text{B + kv}}{\text{g}}$
For the balloon to rise with a constant velocity v, (upward) let the mass be m Here, B - (mg + kv) = 0 …(ii) ⇒ B = mg + kv$\Rightarrow\text{m}=\frac{\text{B}-\text{kw}}{\text{g}}$
So, amount of mass that should be removed = M - m.$=\frac{\text{B + kv}}{\text{g}}-\frac{\text{B}-\text{kv}}{\text{g}}=\frac{\text{B + kv}-\text{B + kv}}{\text{g}}=\frac{2\text{kv}}{\text{g}}\\=\frac{2(\text{Mg}-\text{B})}{\text{G}}=2\Big\{\text{M}-\Big(\frac{\text{B}}{\text{g}}\Big)\Big\}$
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Question 65 Marks
Figure shows a light spring balance connected to two blocks of mass 20kg each. The graduations in the balance measure the tension in the spring.
  1. What is the reading of the balance?
  2. Will the reading change if the balance is heavy, say 2.0kg?
  3. What will happen if the spring is light but the blocks have unequal masses?
Answer
  1.  

The reading of the balance = Tension in the string
And tension in the string = 20g
So, the reading of the balance = 20g = 200N
  1. If the balance is heavy, the reading will not change because the weight of spring balance does not affect the tension in the string.
  2. If the blocks have unequal masses, the spring balance will accelerate towards the heavy block with an acceleration a. Then the reading will be equal to the tension in the string.
Suppose $m_1 > m_2.$
Then tension in the string,
$\text{T}=\frac{2\text{m}_1\text{m}_2\text{g}}{\text{m}_1+\text{m}_2}$
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Question 75 Marks
In the previous problem, suppose $m_2 = 2.0$ kg and $m_3 = 3.0$ kg. What should be the mass m so that it remains at rest?
Answer


$m_1$​​​​​​​ should be at rest.
$\text{T}-\text{m}_1\text{g}=0$
$\Rightarrow\text{T}=\text{m}_1\text{g} \ ...(\text{i})$
$\frac{\text{T}}{2}-2\text{g}-2\text{a}_1=0$
$\Rightarrow\text{T}-4\text{g}-4\text{a}_1=0 \ ...(\text{ii})$
$\frac{\text{T}}{2}-3\text{g}-3\text{a}_1=0$
$\Rightarrow\text{T}=6\text{g}-6\text{a}_1 \ ...(\text{iii})$
From eqn (ii) & (iii) we get
$3\text{T}-12\text{g}=12\text{g}-2\text{T}\Rightarrow\text{T}=\frac{24\text{g}}{5}=408\text{g}.$
Putting yhe value of T eqn (i) we get, $m_1 = 4.8$ kg.
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Question 85 Marks
Consider the Atwood machine of the previous problem. The larger mass is stopped for a moment 2.0s after the system is set into motion. Find the time elapsed before the string is tight again.
Answer
$a = 3.26m/s^2$ $ T = 3.9N$ After 2sec mass
$m_1$ the velocity $V = u + at = 0 + 3.26 × 2 = 6.52$ m/s upward.
At this time $m_2$ is moving 6.52m/s downward. At time 2sec, $m_2$ stops for a moment.
But $m_1$ is moving upward with velocity 6.52m/s.
It will continue to move till final velocity (at highest point) because zero.
Here, $v = 0; u = 6.52 A = -g = -9.8m/s^2$ [moving up ward $m_1$]
$V = u + at$
$⇒ 0 = 6.52 + (-9.8)t$
$\Rightarrow\text{t}=\frac{6.52}{9.8}=0.66=\frac{2}{3}\text{sec}.$
During this period $\frac{2}{3}$sec,
$m_2$ mass also starts moving downward. So the string becomes tight again after a time of $\frac{2}{3}$sec.
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Question 95 Marks
Find the acceleration of the blocks A and B in the three situations shown in figure.


Answer

  1. $5\text{a + T}-5\text{g}=0\Rightarrow\text{T}=5\text{g}-5\text{a} \ ...(\text{i})$ (From FBD - 1)
Again $\Big(\frac{1}{2}\Big)-4\text{g}-8\text{a}=0\Rightarrow\text{T}=8\text{g}-16\text{a} \ ...(\text{ii})$ (From FBD - 2)

From equn (i) and (ii), we get

$\text{5g}-5\text{a}=8\text{g}+16\text{a}\Rightarrow21\text{a}=-3\text{g}\Rightarrow\text{a}=-\frac{1}{7}\text{g}$

So, acceleration of 5kg mass is $\frac{\text{g}}{7}$ upward and that of 4kg mass is $2\text{a}=\frac{2\text{g}}{7}$ (downward).
  1.  


$4\text{a}-\frac{\text{t}}{2}=0\Rightarrow8\text{a}-\text{T}=0\Rightarrow\text{T = 8a} \ ...(\text{ii})$ [from FBD - 4]

Again, $\text{T + 5a}-5\text{g}=0\Rightarrow8\text{a + 5a}-5\text{g}=0$

$\Rightarrow13\text{a}-5\text{g}=0\Rightarrow\text{a =}\frac{5\text{g}}{13}$ downward. (from FBD - 3)

Acceleration of mass
  1. $\text{kg is 2a}=\frac{10}{13}(\text{g}) \ \& \ 5\text{kg}$
  2. is $\frac{5\text{g}}{13}.$
  1.  


$\text{T + 1a}-1\text{g}=0\Rightarrow\text{T = 1g}-1\text{a} \ ...(\text{i})$ [From FBD - 5]

Again, $\frac{\text{T}}{2}-2\text{g}-4\text{a}=0\Rightarrow\text{T}-4\text{g}-8\text{a}=0 \ ...(\text{ii})$ [From FBD - 6]

$\Rightarrow1\text{g}-1\text{a}-4\text{g}-8\text{a}=0$ [From (i)]

$\Rightarrow\text{a}=-\Big(\frac{\text{g}}{3}\Big)$ downward.

Acceleration of mass 1kg(b) is $\frac{\text{g}}{3}\text{(up)}$

Acceleration of mass 2kg(A) is $\frac{2\text{g}}{3}$ (downward).
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Question 105 Marks
Consider the situation shown in figure. Both the pulleys and the string are light and all the surfaces are frictionless.
  1. Find the acceleration of the mass M.
  2. Find the tension in the string.
  3. Calculate the force exerted by the clamp on the pulley A in the figure.
Answer


$\text{Ma}-2\text{T}=0$
$\Rightarrow\text{Ma = 2T}\Rightarrow\text{T} = \frac{\text{Ma}}{2}.$
$\text{T + Ma}-\text{Mg}=0$
$\Rightarrow\frac{\text{Ma}}{2}+\text{ma = Mg}.$ $\Big(\text{because = T}=\frac{\text{Ma}}{2}\Big)$
$\Rightarrow3\text{Ma = 2Mg}\Rightarrow\text{a}=\frac{2\text{g}}{3}$
  1. acceleration of mass M is $\frac{2\text{g}}{3}.$
  2. Tension $\text{T}=\frac{\text{Ma}}{2}=\frac{\text{M}}{2}=\frac{2\text{g}}{3}=\frac{\text{Mg}}{3}$
  3. Let, $R^1 =$ resultant of tensions = force exerted by the clamp on the pulley


$\text{R}^1=\sqrt{\text{T}^2+\text{T}^2}=\sqrt{2}\text{T}$
$\therefore\text{R}=\sqrt{2}\text{T}=\sqrt{2}\frac{\text{Mg}}{3}=\frac{\sqrt{2}\text{Mg}}{3}$
Again, $\tan\theta=\frac{\text{T}}{\text{T}}=1\Rightarrow\theta=45^{\circ}.$
So, it is $\frac{\sqrt{2}\text{Mg}}{3}$ at an angle of 45° with horizontal.
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Question 115 Marks
Consider the situation shown in figure. All the surfaces are frictionless and the string and the pulley are light. Find the magnitude of the acceleration of the two blocks.
Answer

$\sin\theta_1=\frac{4}{5}$
$\sin\theta_2=\frac{3}{5}$
$\text{g}\sin\theta_1-(\text{a + T})=0$
$\Rightarrow\text{g}\sin\theta_1=\text{a + T} \ ...(\text{i})$
$\Rightarrow\text{T + a}-\text{g}\sin\theta_1=0$
$\text{T}-\text{g}\sin\theta_2-\text{a}=0$
$\Rightarrow\text{T = g}\sin\theta_2+\text{a} \ ...(\text{ii})$
$\Rightarrow\text{T + a}-\text{g}\sin\theta_1=0$
From eqn (i) and (ii)
$\text{g}\sin\theta_2+\text{a + a}-\text{g}\sin\theta_1=0$
$\Rightarrow2\text{a = g}\sin\theta_1-\text{g}\sin\theta_2=\text{g}\Big(\frac{4}{5}-\frac{3}{5}\Big)=\frac{\text{g}}{5}$
$\Rightarrow\text{a}=\frac{\text{g}}{5}\times\frac{1}{2}=\frac{\text{g}}{10}$
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Question 125 Marks
A constant force $\text{F}=\frac{\text{m}_2\text{g}}{2}$ is applied on the block of mass $m_1$ as shown in figure. The string and the pulley are light and the surface of the table is smooth. Find the acceleration of $m_1.$​​​​​​​
Answer


From the above Free body diagram
$\text{M}_1\text{a + F}-\text{T}=0\Rightarrow\text{T = m}_1\text{a + F} \ ...(\text{i})$
From the above Free body diagram
$\text{m}_2\text{a + T}-\text{m}_2\text{g}= 0 \ ...(\text{ii})$
$\Rightarrow\text{m}_2\text{a}+\text{m}_1\text{a}+\text{F}-\text{m}_2\text{g}=0$ (from (i))
$\Rightarrow\text{a(m}_1+\text{m}_2)+\frac{\text{m}_2\text{g}}{2}-\text{m}_2\text{g}=0$ $\Big\{\text{because f }=\frac{\text{m}^2\text{g}}{2}\Big\}$
$\Rightarrow\text{a}(\text{m}_1+\text{m}_2)-\text{m}_2\text{g}=0$
$\Rightarrow\text{a}(\text{m}_1+\text{m}_2)=\frac{\text{m}_2\text{g}}{2}\Rightarrow\text{a}\frac{\text{m}_2\text{g}}{2(\text{m}_1=\text{m}_2)}$
Acceleration of mass $m_1$ is $\frac{\text{m}_2\text{g}}{2(\text{m}_1 = \text{ m}_2)}$ towards right.
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Question 135 Marks
Find the acceleration of the block of mass M in the situation shown in figure. All the surfaces are frictionless and the pulleys and the string are light.
Answer

$2\text{Ma + Mg }\sin\theta-\text{T}=0$
$\Rightarrow\text{T = 2Ma + Mg }\sin\theta \ ...(\text{i})$
$2\text{T + 2Ma}-12\text{Mg}=0$
$\Rightarrow2(2\text{Ma + Mg}\sin\theta)+2\text{Ma}-2\text{Mg}=0$ [From (i)]
$\Rightarrow4\text{Ma + 2Mg }\sin\theta+2\text{Ma}-2\text{Mg}=0$
$\Rightarrow6\text{Ma}+2\text{Mg }\sin30^{\circ}-2\text{Mg}=0$
$\Rightarrow6\text{Ma = Mg}\Rightarrow\text{a}=\frac{\text{g}}{6}.$
Acceleration of mass M is $2\text{a = s}\times\frac{\text{g}}{6}=\frac{\text{g}}{3}$ up the plane.
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Question 145 Marks
A pendulum bob of mass 50g is suspended from the ceiling of an elevator. Find the tension in the string if the elevator:
  1. Goes up with acceleration $1.2m/s^2.$
  2. Goes up with deceleration $1.2m/s^2.$
  3. Goes up with uniform velocity.
  4. Goes down with acceleration $1.2m/s^2.$
  5. Goes down with deceleration $1.2m/s^2.$
  6. Goes down with uniform velocity.
Answer
  1. The tension in the string is found out for the different conditions from the free body diagram as shown below.
  2. T - (W + 0.06 × 1.2) = 0
    ⇒ T = 0.05 × 9.8 + 0.05 × 1.2
    = 0.55N.
  3. $\therefore$ T + 0.05 × 1.2 - 0.05 × 9.8 = 0
  4. ⇒ T = 0.05 × 9.8 - 0.05 × 1.2
    = 0.43N.
  5. When the elevator makes uniform motion
  6. T - W = 0
    ⇒ T = W = 0.05 × 9.8
    = 0.49N
  7. T + 0.05 × 1.2 - W = 0
  8. ⇒ T = W - 0.05 × 1.2
    = 0.43N.
  9. T - (W + 0.05 × 1.2) = 0
  10. ⇒ T = W + 0.05 × 1.2
    = 0.55N
  11. When the elevator goes down with uniform velocity acceleration = 0
  12. T - W = 0
    ⇒ T = W = 0.05 × 9.8
    = 0.49N.
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Question 155 Marks
A car is speeding up on a horizontal road with an acceleration a. Consider the following situations in the car.
  1. A ball is suspended from the ceiling through a string and is maintaining a constant angle with the vertical. Find this angle.
  2. A block is kept on a smooth incline and does not slip on the incline. Find the angle of the incline with the horizontal.
Answer
Suppose pendulum makes $\theta$ angle with the vertical. Let, m = mass of the pendulum. From the free body diagram
$\text{T}\cos\theta-\text{mg}=0$
$\Rightarrow\text{T}\cos\theta=\text{mg}$
$\Rightarrow\text{T}=\frac{\text{mg}}{\cos\theta} \ ..(\text{i})$

$\text{ma}-\text{T}\sin\theta=0$
$\Rightarrow\text{ma = T}\sin\theta$
$\Rightarrow\text{t}=\frac{\text{ma}}{\sin\theta}\ ...(\text{ii})$
From (i) and (ii) $\frac{\text{mg}}{\cos\theta}=\frac{\text{ma}}{\sin\theta}\Rightarrow\tan\theta=\frac{\text{a}}{\text{g}}\Rightarrow\theta=\tan^{-1}\frac{\text{a}}{\text{g}}$
The angle is $\tan^{-1}\Big(\frac{\text{a}}{\text{g}}\Big)$ with vertical. m → mass of block. Suppose the angle of incline is $'\theta'$
From the diagram$\text{ma}\cos\theta-\text{mg}\sin\theta=0$
$\Rightarrow\text{ma}\cos\theta=\text{mg}\sin\theta\Rightarrow\frac{\sin\theta}{\cos\theta}=\frac{\text{a}}{\text{g}}$
$\Rightarrow\tan\theta=\frac{\text{a}}{\text{g}}\Rightarrow\theta=\tan^{-1}\Big(\frac{\text{a}}{\text{g}}\Big).$
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Question 165 Marks
In a simple Atwood machine, two unequal masses $m_1$ and $m_2$ are connected by a string going over a clamped light smooth pulley. In a typical arrangement $m_1 = 300g$ and $m_2 = 600g$. The system is released from rest.
  1. Find the distance travelled by the first block in the first two seconds.
  2. Find the tension in the string.
  3. Find the force exerted by the clamp on the pulley.
Answer


$m_1 = 0.3kg, m_2 = 0.6kg$
$T - (m_1g + m_1a) = 0 …(i)$
$\Rightarrow T = m_1g + m_1a$
$T + m_2a - m_2g = 0 …(ii)$
$\Rightarrow T = m_2g - m_2a$
From equation (i) and equation (ii)
$m_1g + m_1a + m_2a - m_2g = 0, from (i)$
$\Rightarrow a(m_1 + m_2) = g(m_2 - m_1)$
$\Rightarrow\text{a = f}\Big(\frac{\text{m}_2-\text{m}_1}{\text{m}_1-\text{m}_2}\Big)=9.8\Big(\frac{0.6-0.3}{0.6+0.3}\Big)=3.266\text{ms}^{-2}.$
  1. t = 2 sec acceleration $ = 3.266 ms^{-2}$​​​​​​​
Initial velocity u = 0
So, distance travelled by the body is,
$\text{S = ut}+\frac{1}{2}\text{at}^2\Rightarrow0+\frac{1}{2}(3.266)2^2=6.5\text{m}$
  1. From $(i) T = m_1(g + a) = 0.3(9.8 + 3.26) = 3.9N$
  2. The force exerted by the clamp on the pully is given by
F - 2T = 0
F = 2T = 2 × 3.9 = 7.8N.
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Question 175 Marks
Let $m_1 = 1kg, m_2 = 2$ kg and $m_3 = 3$ kg in figure. Find the accelerations of $m_1, m_2$ and $m_3$. The string from the upper pulley to $m_1$ is 20cm when the system is released from rest. How long will it take before m, strikes the pulley?
Answer

Let the block $m+1+$ moves upward with acceleration a, and the two blocks $m_2 $ an $m_3$ have relative acceleration $a_2$  due to the difference of weight between them. So, the actual acceleration at the blocks $m_1, m_2$​​​​​​​  and $m_3 $ will be $a_1.$ 
$(a_1 - a_2)$ and $(a_1 + a_2) $ as shown
$T = 1g - 1a_2 = 0 ...(i)$ from fig. (2)
$\frac{\text{T}}{2}-2\text{g}-2(\text{a}_1-\text{a}_2)=0 \ ...(\text{ii})$ from fig. (3)
$\frac{\text{T}}{2}-3\text{g}-3(\text{a}_1+\text{a}_2)=0 \ ...(\text{iii})$ from fig. (4)
From eqn (i) and eqn (ii), eliminating T
we get, $1g + 1a_2 = 4g + 4(a_1 + a_2) $
$\Rightarrow 5a_2 - 4a_1 = 3g (iv)$
From eqn (ii) and eqn (iii),
we get $2g + 2(a_1 - a_2) = 3g - 3(a_1 - a_2) $
​​​​​​​$\Rightarrow 5a_1 + a_2 = (v)$
Solving (iv) and (v) $\text{a}_1=\frac{2\text{g}}{29}$ and $\text{a}_2=\text{g}-5\text{a}_1=\text{g}-\frac{10\text{g}}{29}=\frac{19\text{g}}{29}$
So, $\text{a}_1-\text{a}_2=\frac{2\text{g}}{29}-\frac{19\text{g}}{29}=-\frac{17\text{g}}{29}$
$\text{a}_1+\text{a}_2=\frac{2\text{g}}{29}+\frac{19\text{g}}{29}=\frac{21\text{g}}{29}$ acceleration of $m_1, m_2, m_3$  ae $\frac{19\text{g}}{29}\text{(up)}\frac{17\text{g}}{29}(\text{down})\frac{21\text{g}}{29}(\text{down})$ respectively.
Again, for $m_1, u = 0, s = 20cm = 0.2m$ and $\text{a}_2=\frac{19}{29}\text{g}$ [g = 10m/s^2]
$\therefore\text{S = ut}+\frac{1}{2}\text{at}^2=0.2=\frac{1}{2}\times\frac{19}{29}\text{gt}^2\Rightarrow\text{t}=0.25\sec.$
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Question 185 Marks
Figure shows a man of mass 60kg standing on a light weighing machine kept in a box of mass 30kg. The box is hanging from a pulley fixed to the ceiling through a light rope, the other end of which is held by the man himself. If the man manages to keep the box at rest, what is the weight shown by the machine? What force should he exert on the rope to get his correct weight on the machine?
Answer

  1. Given, Mass of man = 60kg.
Let R' = apparent weight of man in this case.
Now, R' + T - 60g = 0 [From FBD of man]
⇒ T = 60g - R' ...(i)
T - R' - 30g = 0 ...(ii) [From FBD of box]
⇒ 60g - R' - R' - 30g = 0 [From (i)]
⇒ R' = 15g The weight shown by the machine is 15kg.
  1. To get his correct weight suppose the applied force is ‘T’ and so, acclerates upward with ‘a’. In this case, given that correct weight = R = 60g, where g = 10 due to gravity

From the FBD of the man
$ T^1 + R - 60g - 60a = 0$
$\Rightarrow T^1 - 60a = 0 [\therefore R = 60g]$
$\Rightarrow T^1 = 60a ...(i)$​​​​​​​

From the FBD of the box
$T_1 - R - 30g - 30a = 0$
$\Rightarrow T^1 - 60g - 30g - 30a = 0$
$\Rightarrow T^1 - 30a = 90g = 900$
$\Rightarrow T^1 = 30a - 900 ...(ii)$
From eqn (i) and eqn (ii) we get
$T^1 = 2T^1 - 1800 \Rightarrow T^1 = 1800N.$
$\therefore$ So, he should exert 1800N force on the rope to get correct reading.
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Question 195 Marks
Find the reading of the spring balance shown in figure. The elevator is going up with an acceleration of $\frac{\text{g}}{10},$ the pulley and the string are light and the pulley is smooth.
Answer
Let the acceleration of the 3kg mass relative to the elevator is ‘a’ in the downward direction.
As, shown in the free body diagram$\text{T}-1.5\text{g}-1.5\Big(\frac{\text{g}}{10}\Big)-1.5\text{a}=0$ from figure (1)
and, $\text{T}-3\text{g}-3\Big(\frac{\text{g}}{10}\Big)+3\text{a}=0$ from figure (2)$\Rightarrow\text{T}=1.5\text{g}+1.5\Big(\frac{\text{g}}{10}\Big)+1.5\text{a}\ ...(\text{i})$
And $\text{T}=3\text{g}+3\Big(\frac{\text{g}}{10}\Big)-3\text{a} \ ...(\text{ii})$ Equation (i) × 2 $\Rightarrow3\text{g}+3\Big(\frac{\text{g}}{10}\Big)+3\text{a}=2\text{T}$ Equation (ii) × 1 $\Rightarrow3\text{g}+3\Big(\frac{\text{g}}{10}\Big)-3\text{a}=\text{T}$ Subtracting the above two equations we get, T = 6a Subtracting T = 6a in equation (ii)$6\text{a}=3\text{g}+3\Big(\frac{\text{g}}{10}\Big)-3\text{a}$
$\Rightarrow9\text{a}=\frac{33\text{g}}{10}\Rightarrow\text{a}=\frac{(9.8)33}{10}=32.34$
$\Rightarrow\text{a}=3.59$
$\therefore\text{T}=6\text{a}=6\times3.59=21.55$
$\text{T}^1=2\text{T}=2\times21.55=43.1\text{N}$ cut is $T_1$ shown in spring.
$\text{Mass}=\frac{\text{wt}}{\text{g}}=\frac{43.1}{9.8}=4.39=4.4\text{kg}$
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Question 205 Marks
In figure $m_1 = 5kg, m_1 = 2kg$ and F = 1N. Find the acceleration of either block. Describe the motion of $m_1$ if the string breaks but F continues to act.
Answer

From the above free body diagram
$T + m_1a - m(m_1g + F ) = 0$
From the free body diagram
$T - (m_2g + F + m_2a) = 0$
$\Rightarrow T = m_1g + F - m_1a$
$\Rightarrow T = 5g + 1 - 5a …(i)$
$\Rightarrow T = m_2g + F + m_2a$
$\Rightarrow T = 2g + 1 + 2a …(ii)$
From the eqn (i) and eqn (ii) $5g + 1 - 5a = 2g + 1 + 2a$
$⇒ 3g - 7a = 0 $
$⇒ 7a = 3g$
$\Rightarrow\text{a}=\frac{3\text{g}}{7}=\frac{29.4}{7}=4.2\text{m/s}^2$
​​​​​​​$[g = 9.8m/s^2]$​​​​​​​
  1. Acceleration of block is $4.2m/s^2$​​​​​​​
  2. After the string breaks $m_1$ move downward with force F acting down ward.
$m_1a = F + m_1g = (1 + 5g) = 5(g + 0.2)$
Force = 1N, acceleration $=\frac{1}{5}=0.2\text{m/s.}$
So, acceleration $=\frac{\text{Force}}{\text{mass}}=\frac{5(\text{g}+0.2)}{5}=(\text{g}+0.2)\text{m/s}^2$
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Question 215 Marks
Find the mass M of the hanging block in figure which will prevent the smaller block from slipping over the triangular block. All the surfaces are frictionless and the strings and the pulleys are light.
Answer

As the block ‘m’ does not slinover M', ct will have same acceleration as that of M' From the freebody diagrams.
$\text{T + Ma}-\text{Mg}=0 \ ...(\text{i})$ (From FBD - 1)
$\text{T}-\text{M}'\text{a}-\text{R}\sin\theta=0 \ ...(\text{ii})$ (From FBD - 2)
$\text{R}\sin\theta-\text{ma}=0 \ ...(\text{iii})$ (From FBD - 3)
$\text{R}\cos\theta-\text{mg}=0 \ ...(\text{iv})$ (From FBD - 4)
Eliminating T, R and a from the above equation, we get $\text{M}=\frac{\text{M}'+\text{m}}{\cot\theta-1}$
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Question 225 Marks
Find the acceleration of the 500g block in figure.
Answer


$m_1 = 100g = 0.1kg$
$m_2 = 500g = 0.5kg$
$m_3 = 50g = 0.05kg.$
$T + 0.5a - 0.5g = 0 ...(i)$
$T_1- 0.5a - 0.05g = a ...(ii)$
$T_1 + 0.1a - T + 0.05g = 0 ...(iii)$
From equn $(ii) T_1 = 0.05g + 0.05a ...(iv)$
From equn $(i) T_1 = 0.5g - 0.5a ...(v)$
Equn (iii) becomes $T_1 + 0.1a - T + 0.05g = 0​​​​​​​$​​​​​​​

$⇒ 0.05g + 0.05a + 0.1a - 0.5g + 0.5a + 0.05g = 0 [$From $(iv)$ and $(v)]$
$\Rightarrow0.65\text{a}=0.4\text{g}$
$\Rightarrow\text{a}=\frac{0.4}{0.65}=\frac{40}{65}\text{g}=\frac{8}{13}\text{g}$ downward
Acceleration of 500gm block is $\frac{8\text{g}}{13\text{g}}$ downward
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MCQ 235 Marks
The force exerted by the floor of an elevator on the foot of a person standing there is more than the weight of the person if the elevator is:
  • A
    Going up and slowing down.
     
  • B
    Going up and speeding up.
     
  • C
    Going down and slowing down.
     
  • D
    Going down and speeding up.
Answer
  1. Going up and speeding up.
  2. Going down and slowing down.
Explanation:
It means normal force exerted by the floor of the elevator on the person is greater that the weight of the person.
i.e. N > mg
  1. Going up and speeding up:
$a_{eff} = g + a$
$N = ma_{eff} = mg + ma (N > mg)$
  1. Going down and speeding up:
$a_{eff} = g - a$
$N = mg - ma (N < mg)$​​​​​​​​​​​​​​
  1. Going down and slowing down:
$a_{eff} = g - (-a) = g + a$
$N = mg + ma (N > mg)$​​​​​​​​​​​​​​
  1. Going up and slowing down:
$a_{eff} = g - a$
$N = mg - ma (N < mg)$
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MCQ 245 Marks
A block of mass $10\ kg$ is suspended through two light spring balances as shown in figure:
  • ✓
    Both the scales will read $10\ kg.$
  • B
    Both the scales will read $5\ kg.$
  • C
    The upper scale will read $10\ kg$ and the lower zero.
  • D
    The readings may be anything but their sum will he $10\ kg.$
Answer
Correct option: A.
Both the scales will read $10\ kg.$

From the free$-$body diagram,
$K_1x_1 = mg = 10 \times 9.8 = 98N$
$K_2x_2 = K_1x_1$
So, $ K_1x_1 = K_2x_2 = 98N$
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MCQ 255 Marks
Three rigid rods are joined to form an equilateral triangle $\text{ABC}$ of side $1m$. Three particles carrying charges $20\mu\text{C}$ each are attached to the vertices of the triangle. The whole system is at rest in an inertial frame. The resultant force on the charged particle at $A$ has the magnitude:
  • ✓
    Zero
  • B
    $3.6N$
  • C
    $3.6\sqrt{3}\text{N}$
  • D
    $7.2N.$
Answer
Correct option: A.
Zero
Using, $F_\text{net} = ma,$
$a = 0 $
$\Rightarrow F_\text{net} {= 0}$
As the whole system is at rest, the resultant force on the charged particle at $A$ is zero.
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MCQ 265 Marks
A force $F_1$ acts on a particle so as to accelerate it from rest to $a$ velocity $v$. The force $F_1$ is then replaced by $F_2$ which decelerates it to rest:
  • A
    $F_1$ must be equal to $F_2.$
  • ✓
    $F_1$ may be equal to $F_2.$
  • C
    $F_1$ must be unequal to $F_2.$
  • D
    None of these.
Answer
Correct option: B.
$F_1$ may be equal to $F_2.$
Any force applied in the direction opposite the motion of the particle decelerates it to rest.
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MCQ 275 Marks
A free $^{238}U$ nucleus kept in a train emits an alpha particle. When the train is stationary, a nucleus decays and a passenger measures that the separation between the alpha particle and the recoiling nucleus becomes x at time t after the decay. If the decay takes place while the train is moving at a uniform velocity v, the distance between the alpha particle and the recoiling nucleus at a time t after the decay as measured by the passenger is:
  • A
    x + vt
     
  • B
    x - vt
     
  • ✓
    x
     
  • D
    Depends on the direction of the train.
Answer
Correct option: C.
x
 
  1. x
Explanation:

The moving train does not put any extra force on the alpha particle and the recoiling nucleus. So, the distance between the alpha particle and the recoiling nucleus at a time tafter the decay, as measured by the passenger, will be same as before, i.e. x.
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Question 285 Marks
A block of mass m is placed on a smooth inclined plane of inclination 0 with the horizontal. The force exerted by the plane on the block has a magnitude:
  1. $\text{mg}$
  2. $\frac{\text{mg}}{\cos\theta}$
  3. $\text{mg}\cos\theta$
  4. $\text{mg}\tan\theta$
Answer
  1. $\text{mg}\cos\theta$
Explanation:



From the free-body diagram,

$\text{N = mg}\cos\theta$

Normal force exerted by the plane on the block is $\text{mg}\cos\theta.$
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Question 295 Marks
Neglect the effect of rotation of the earth. Suppose the earth suddenly stops attracting objects placed near its surface. A person standing on the surface of the earth will:
  1. Fly up.
  2. Slip along the surface.
  3. Fly along a tangent to the earth's surface.
  4. Remain standing.
Answer
  1. Remain standing.
Explanation:

If the earth suddenly stops attracting objects placed near its surface, the net force on the person will become zero and according to the first law of motion, the person will remain standing.
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Question 305 Marks
A smooth wedge A is fitted in a chamber hanging from a fixed ceiling near the earth's surface. A block B placed at the top of the wedge takes a time T to slide down the length of the wedge. If the block is placed at the top of the wedge and the cable supporting the chamber is broken at the same instant, the block will:
  1. Take a time longer than T to slide down the wedge.
  2. Take a time shorter than T to slide down the wedge.
  3. Remain at the top of the wedge.
  4. Jump off the wedge.
Answer
  1. Remain at the top of the wedge.
Explanation:

Downward gravitational force will be balanced by the upward pseudo force (because of the motion of the wedge in downward direction). The block will remain at its position, as both the box and the inclined plane are falling with the same acceleration (g).
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MCQ 315 Marks
In an imaginary atmosphere, the air exerts a small force $F$ on any particle in the direction of the particle's motion. A particle of mass $m$ projected upward takes a time $t_1$ in reaching the maximum height and $t_2$ in the return journey to the original point. Then:
  • A
    $t_1 < t_2$
  • ✓
    $t_1 > t_2$
  • C
    $t_1 = t_2$
  • D
    The relation between $t_1$ and $t_2$ depends on the mass of the particle.
Answer
Correct option: B.
$t_1 > t_2$
Let acceleration due to air resistance force be $a$.
Let $H$ be maximum height attained by the particle.
Direction of air resistance force is in the direction of motion.
In the upward direction of motion, $\text{a}_{\text{eff}}=|\text{g}-\text{a}|.$
$\text{t}_{1}=\sqrt{\frac{2\text{H}}{|\text{g}-\text{a}|}} \ ...(1)$
In the downward direction of motion, $\text{a}_{\text{eff}}=\text{g}+\text{a}.$
$\text{t}_{2}=\sqrt{\frac{2\text{H}}{\text{g}+\text{a}}} \ ...(2)$
So, $t_1 > t_2$
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MCQ 325 Marks
A person standing on the floor of an elevator drops a coin. The coin reaches the floor of the elevator in a time $t_1$ if the elevator is stationary and in time $t_2$ if it is moving uniformly. Then :
  • ✓
    $t_1 = t_2$
  • B
    $t_1 < t_2$
  • C
    $t_1 > t_2$
  • D
    $t_1 < t_2$ or $t_1 > t_2$ depending on whether the lift is going up or down.
Answer
Correct option: A.
$t_1 = t_2$
After the coin is dropped, the only force acting on it is gravity, which is same for both the cases.
So $t_1= t_2$
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Question 335 Marks
A person says that he measured the acceleration of a particle to be non-zero while no force was acting on the particle:
  1. He is a liar.
  2. His clock might have run slow.
  3. His meter scale might have been longer than the standard.
  4. He might have used non-inertial frame.
Answer
  1. He might have used non-inertial frame.
Explanation:

If no force is acting on a particle and yet, its acceleration is non-zero, it means the observer is in a non-inertial frame.
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Question 345 Marks
A reference frame attached to the earth:
  1. Is an inertial frame by definition.
  2. Cannot be an inertial frame because the earth is revolving around the sun.
  3. Is an inertial frame because Newton's laws are applicable in this frame.
  4. Cannot be an inertial frame because the earth is rotating about its axis.
Answer
  1. Cannot be an inertial frame because the earth is revolving around the sun.
  1. Cannot be an inertial frame because the earth is rotating about its axis.
Explanation:

A reference frame attached to the earth cannot be an inertial frame because the earth is revolving around the sun and also rotating about its axis.
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MCQ 355 Marks
A particle is observed from two frames $S_1$ and. $S_2$. The frame $S_2$ moves with respect to $S_1$ with an acceleration a. Let $F_1$ and $F_2$ be the pseudo forces on the particle when seen from $S_1$ and $S_2$ respectively. Which of the following are not possible?
  • A
    $F_1=0, F_2 \neq 0$
  • B
    $F_1 \neq 0, F_2=0$
  • C
    $F_1 \neq 0, F_2 \neq 0$
  • ✓
    $F_1=0, F_2=0$
Answer
Correct option: D.
$F_1=0, F_2=0$
$\text{a}_{\text{s}_1\text{s}_2}=\text{a} \ ...(1) $
Acceleration of the particle $\text{w.r.t}$. to $\text{S}_1=\frac{\text{F}_1}{\text{m}}$
Acceleration of the particle $\text{w.r.t}$. to $\text{S}_2=\frac{\text{F}_2}{\text{m}}$
If we assume $F_1 = 0$ and $F_2 = 0,$
we can conclude that $\text{a}_{\text{s}_2\text{s}_1}=0 \ ...(2)$
From equations $(1)$ and $(2), $we can say that our assumption is wrong.
And $F_1=0, F_2=0$ is not possible.
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MCQ 365 Marks
Figure shows a heavy block kept on a frictionless surface and being pulled by two ropes of equal mass $m$. At $t = 0,$ the force on the left rope is withdrawn but the force on the right end continues to act. Let $F_1$ and $F_2$ be the magnitudes of the forces by the right rope and the left rope on the block respectively:
  • ✓
    $F_1= F_2 = F$ for $t < 0.$
  • B
    $F_1 = F_2 = F + mg$ for $t < 0.$
  • C
    $F_1= F, F_2 = F$ for $t > 0.$
  • D
    $F_1< F, F_2 = F$ for $t > 0.$
Answer
Correct option: A.
$F_1= F_2 = F$ for $t < 0.$
At $t < 0,$ the block is in equilibrium in the horizontal direction.
So, $F_1= F_2 = F$
At $t > 0, F_2= 0$ and $F_1= F.$
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Question 375 Marks
A particle stays at rest as seen in a frame. We can conclude that:
  1. The frame is inertial.
  2. Resultant force on the particle is zero.
  3. The frame may be inertial but the resultant force on the particle is zero.
  4. The frame may be non-inertial but there is a non-zero resultant force.
Answer
  1. The frame may be inertial but the resultant force on the particle is zero.
  2. The frame may be non-inertial but there is a non-zero resultant force.
Explanation:

According to Newton's second law which says that net force acting on the particle is equal to rate of change of momentum (or mathematically F = ma), so if a particle is at rest then $\text{F}_{\text{net}}=\text{ma = m}\frac{\text{dv}}{\text{dt}}=\text{m}\frac{\text{d}(0)}{\text{dt}}=\text{m}\times0=0.$

Now, if the frame is inertial, then the resultant force on the particle is zero.

If the frame is non-inertial, vector sum of all the forces plus a pseudo force is zero.

i.e $\text{F}_{\text{net}}\neq0.$
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Question 385 Marks
A car accelerates on a horizontal road due to the force exerted by:
  1. The engine of the car.
  2. The driver of the car.
  3. The earth.
  4. The road.
Answer
  1. The road.
Explanation:

The car pushes the ground in the backward direction and according to the third law of motion, reaction force of the ground in the forward direction acts on the car.
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Question 395 Marks
When a horse pulls a cart, the force that helps the horse to move forward is the force exerted by:
  1. The cart on the horse.
  2. The ground on the horse.
  3. The ground on the cart.
  4. The horse on the ground.
Answer
  1. The ground on the horse.
Explanation:

The horse pushes the ground in the backward direction and, in turn, the ground pushes the horse in the forward direction, according to Newton's third law of motion.
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Question 405 Marks
If the tension in the cable supporting an elevator is equal to the weight of the elevator, the elevator may be:
  1. Going up with increasing speed.
  2. Going down with increasing speed.
  3. Going up with uniform speed.
  4. Going down with uniform speed.
Answer
  1. Going up with uniform speed.
  2. Going down with uniform speed.
Explanation:

Tension in the cable = Weight of the elevator

Or, total upward force = total downward force

That is, there's no acceleration or uniform velocity.

So, the elevator is going up/ down with uniform speed.
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Question 415 Marks
A block of mass m is placed on a smooth wedge of inclination 0. The whole system is accelerated horizontally so that the block does not slip on the wedge. The force exerted by the wedge on the block has a magnitude:
  1. $\text{mg}$
  2. $\frac{\text{mg}}{\cos\theta}$
  3. $\text{mg}\cos\theta$
  4. $\text{mg}\tan\theta$
Answer
  1. $\frac{\text{mg}}{\cos\theta}$
Explanation:



Free-body Diagram of the Small Block of Mass 'm'

The block is at equilibrium w.r.t. to wedge. Therefore,

$\text{mg}\sin\theta=\text{ma}\cos\theta$

$\Rightarrow\text{a}=\text{g}\tan\theta$

Normal reaction on the block is

$\text{N = mg}\cos\theta+\text{ma}\sin\theta$

Putting the value of a, we get:

$\text{N = mg}\cos\theta+\text{mg}\tan\theta\sin\theta$

$\text{N = mg}\cos\theta+\text{mg}\frac{\sin\theta}{\cos\theta}\sin\theta$

$\text{N}=\frac{\text{mg}}{\cos\theta}$
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Question 425 Marks
Two objects A and B are thrown upward simultaneously with the same speed. The mass of A is greater than the mass of B. Suppose the air exerts a constant and equal force of resistance on the two bodies:
  1. The two bodies will reach the same height.
  2. A will go higher than B.
  3. B will go higher than A.
  4. Any of the above three may happen depending on the speed with which the objects are thrown.
Answer
  1. A will go higher than B.
Explanation:

Let the air exert a constant resistance force = F (in downward direction).

Acceleration of particle A in downward direction due to air resistance, $\text{a}_{\text{A}}=\frac{\text{F}}{\text{m}_{\text{A}}}.$

Acceleration of particle B in downward direction due to air resistance, $\text{a}_{\text{B}}=\frac{\text{F}}{\text{m}_{\text{B}}}.$

$\text{m}_{\text{A}}>\text{m}_{\text{B}}$

$\text{a}_{\text{A}}<\text{a}_{\text{B}}$

$\text{S = us}+\frac{1}2{}\text{at}^2$

So, $\text{H}_{\text{A}}=\text{ut}-\frac{1}{2}\big(\text{a}_{\text{A}}+\text{g}\big)\text{t}^2$

$\text{H}_{\text{B}}=\text{ut}-\frac{1}{2}\big(\text{a}_{\text{B}}+\text{g}\big)\text{t}^2$

$\text{H}_{\text{A}}>\text{H}_{\text{B}}$

Therefore, A will go higher than B.
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MCQ 435 Marks
A particle is found to be at rest when seen from a frame $S_1$ and moving with constant velocity when seen from another frame $S_2$. Mark out the possible options:
  • A
    Both the frames are inertial.
     
  • B
    Both the frames are non-inertial.
     
  • C
    $S_1$ is inertial and $S_2$ is non-inertial.
     
  • D
    $S_1$ is non-inertial and $S_2$ is inertial.
Answer
  1. Both the frames are inertial.
  2. Both the frames are non-inertial.
Explanation:
$S_1$ is moving with constant velocity w.r.t frame $S_2$. So, if $S_1$ is inertial, then $S_2$ will be inertial and if $S_1$ is non-inertial, then $S_2$ will be non-inertial.
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Question 445 Marks
Figure shows the displacement of a particle going along the X-axis as a function of time. The force acting on the particle is zero in the region:
  1. AB
  2. BC
  3. CD
  4. DE
Answer
  1. AB
  1. CD
Explanation:

Slope of the x-t graph gives velocity. In the regions AB and CD, slope or velocity is constant, i.e. acceleration is zero. Hence, from the second law, force is zero in these regions.
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MCQ 455 Marks
A body of weight $w_1$ is suspended from the ceiling of a room through a chain of weight $w_2$. The ceiling pulls the chain by a force:
  • A
    $w_1$
  • B
    $w_2$
  • ✓
    $w_1 + w_2$
  • D
    $\frac{\text{w}_1+\text{w}_2}{2}$
Answer
Correct option: C.
$w_1 + w_2$

From the free $-$ body diagram,
$(w_1 + w_2) - N = 0$
$N = w_1 + w_2$
The ceiling pulls the chain by a force $(w_1 + w_2).$
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