5 Marks Questions - Physics STD 12 Science Questions - Vidyadip

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22 questions · self-marked practice — reveal the answer and mark yourself.

Question 15 Marks

Suppose the block of the previous problem is pushed down the incline with a force of 4N. How far will the block move in the first two seconds after starting from rest? The mass of the block is 4kg.

Answer

From the free body diagram
$4-4\text{a}-\mu\text{R}+4\text{g}\sin30^{\circ}=0 \ ...(1)$
$\text{R}-4\text{g}\cos30^{\circ}=0 \ ...(2)$
$\Rightarrow\text{R}=4\text{g}\cos30^{\circ}$
Putting the values of R is & in equn. (1)
$4-4\text{a}-0.11\times4\text{g}\cos30^{\circ}+4\text{g}\sin30^{\circ}=0$
$\Rightarrow4-4\text{a}-0.11\times4\times10\times\Big(\frac{\sqrt{3}}{2}\Big)\\+4\times10\times\Big(\frac{1}{2}\Big)=0$
$\Rightarrow4-4\text{a}-3.81+20=0\Rightarrow\text{a}\approx5\text{m/s}^2$
For the block $\text{u}=0,\text{t}=2\text{sec},\text{a}=5\text{m/s}^2$
Distance $\text{s = ut}+\frac{1}{2}\text{at}^2\Rightarrow\text{s}=0+\Big(\frac{1}{2}\Big)5\times2^2=10\text{m}$
The block will move 10m.

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Question 25 Marks

Consider the situation shown in figure. Suppose a small electric field E exists in the space in the vertically upward direction and the upper block carries a positive charge Q on its top surface. The friction coefficient between the two blocks is g but the floor is smooth. What maximum horizontal force F can be applied without disturbing the equilibrium?
[Hint: The force on a charge Q by the electric field E is F = QE in the direction of E.]

Answer

$\text{R}_1+\text{QE}-\text{mg}=0$
$\text{R}_1=\text{mg}-\text{QE} \ ...(\text{i})$
$\text{F}-\text{T}-\mu\text{R}_1=0$
$\Rightarrow\text{F}-\text{T}=\mu(\text{mg}-\text{QE}) = 0$
$\Rightarrow\text{F}-\text{T}=\mu\text{mg}+\mu\text{QE}=0 \ ...(\text{ii})$
$\text{T}-\mu\text{R}_1=0$
$\Rightarrow\text{T}=\mu\text{R}_1=\mu(\text{mg}-\text{QE})=\mu\text{mg}-\mu\text{QE}$
Now equation (ii) is $\text{F}-\text{mg}+\mu\text{QE}-\mu\text{mg}+\mu\text{QE}=0$
$\Rightarrow\text{F}-2\mu\text{mg}+2\mu\text{QE}=0$
$\Rightarrow\text{F}=2\mu\text{mg}-2\mu\text{QE}$
$\Rightarrow\text{F}=2\mu(\text{mg}-\text{QE})$
Maximum horizontal force that can be applied is $2\mu(\text{mg}-\text{QE}).$

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Question 35 Marks

A block of mass M is kept on a rough horizontal surface. The coefficient of static friction between the block and the surface is $\mu.$ The block is to be pulled by applying a force to it. What minimum force is needed to slide the block? In which direction should this force act?

Answer

Let ‘p’ be the force applied to at an angle $\theta$
From the free body diagram
$\text{R + P}\sin\theta-\text{mg}=0$
$\Rightarrow\text{R}=-\text{P}\sin\theta+\text{mg} \ ...(\text{i})$
$\mu\text{R}-\text{p}\cos\theta \ ...(\text{ii})$
Equn. (i) is
$\mu(\text{mg}-\text{P}\sin\theta)-\text{P}\cos\theta=0$
$\Rightarrow\mu\text{mg}=\mu\rho\sin\theta-\text{P}\cos\theta\rightarrow\rho=\frac{\mu\text{mg}}{\mu\sin\theta+\cos\theta}$
Applied force P should be minimum, when $\mu\sin\theta+\cos\theta$ is maximum.
Again, $\mu\sin\theta+\cos\theta$ is maximum when its derivative is zero.
$\therefore\frac{\text{d}}{\text{d}\theta}(\mu\sin\theta+\cos\theta)=0$
$\Rightarrow\mu\cos\theta-\sin\theta=0\Rightarrow\theta\tan^{-1}\mu$
So, $\text{p}=\frac{\mu\text{mg}}{\mu\sin\theta+\cos\theta}=\frac{\frac{\mu\text{mg}}{\cos\theta}}{\frac{\mu\sin\theta}{\cos\theta}+\frac{\cos\theta}{\cos\theta}}$
$=\frac{\mu\text{mg}\sec\theta}{1+\mu\tan\theta}=\frac{\mu\text{mg}\sec\theta}{1+\tan^2\theta}$
$=\frac{\mu\text{mg}}{\sec\theta}=\frac{\mu\text{mg}}{\sqrt{(1+\tan^2\theta)}}=\frac{\mu\text{mg}}{\sqrt{1+\mu^2}}$
Minimum force is $\frac{\mu\text{mg}}{\sqrt{1+\mu^2}}$ at an angle $\theta=\tan^{-1}\mu.$

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Question 45 Marks

Find the accelerations $a_1, a_2, a_3$ of the three blocks shown in figure if a horizontal force of $10N$ is applied on:

$2\ kg$ block.
$3\ kg$ block.
$7\ kg$ block. Take $g = 10m/s^2.$

Answer

When the $10N $force applied on $2\ kg$ block, it experiences maximum frictional force $\mu\text{R}_1=\mu\times2\text{kg}=(0.2)\times20=4\text{N}$ from the $3\ kg$ block.
So, the $2\ kg$ block experiences a net force of $10 - 4 = 6N$
So, $\text{a}_1=\frac{6}{2}=3\text{m/s}^2.$
But for the $3\ kg$ block, $($fig$-3)$ the frictional force from $2\ kg$ block $(4N)$ becomes the driving force and the maximum frictional force between $3\ kg$ and $7\ kg$ block is
$\mu_2\text{R}_2=(0.3)\times5\text{kg}=15\text{N}$
So, the $3\ kg$ block cannot move relative to the $7\ kg$ block. The $3\ kg$ block and $7\ kg$ block both will have same acceleration $(a_2 = a_3)$ which will be due to the $4N$ force because there is no friction from the floor.
$\therefore\text{a}_2=\text{a}_3=\frac{4}{10}=0.4\text{m/s}^2$

When the $10N$ force is applied to the $3\ kg$ block, it can experience maximum frictional force of $15 + 4 = 19N$ from the $2\ kg$ block $7\ kg$ block.
So, it can not move with respect to them.
As the floor is frictionless, all the three bodies will move together
$\therefore\text{a}_1=\text{a}_2=\text{a}_3=\frac{10}{12}=\Big(\frac{5}{6}\Big)\text{m/s}^2$
Similarly, it can be proved that when the $10N$ force is applied to the $7\ kg$ block, all the three blocks will move together.
Again $\text{a}_1=\text{a}_2=\text{a}_3=\Big(\frac{5}{6}\Big)\text{m/s}^2$

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Question 55 Marks

The friction coefficient between the table and the block shown in figure is 0.2. Find the tensions in the two strings.

Answer

From the free body diagram
$\text{T + 15a}-15\text{g}=0$
$\Rightarrow\text{T = 15g}-15\text{a} \ ...(1)$
$\text{T}-\text{(T}_1+5\text{a}+\mu\text{R})=0$
$\Rightarrow\text{T}-(5\text{g + 5a + 5a}+\mu\text{R})=0$
$\Rightarrow\text{T = 5g + 10a}+\mu\text{R} \ ...(\text{ii})$
$\text{T}_1-5\text{g}-5\text{a}=0$
$\Rightarrow\text{T}_1=5\text{g + 5a} \ ...(\text{iii})$
From (i) & (ii)
$15\text{g}-15\text{a = 5g + 10a + 0.2 (5g)}$
$\Rightarrow25\text{a}=90\Rightarrow\text{a}=3.6\text{m/s}^2$
Equation (ii)
$\Rightarrow\text{T}=5\times10+10\times3.6+0.2\times5\times10$
$\Rightarrow96\text{N}$ in the left string
Equation (iii) $\text{T}_1=5\text{g + 5a}=5\times10+5\times3.6=68\text{N}$ in the right string.

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Question 65 Marks

Two masses $M_1$ and $M_2$ are connected by a light rod and the system is slipping down a rough incline of angle $\theta$ with the horizontal. The friction coefficient at both the contacts is $\mu.$ Find the acceleration of the system and the force by the rod on one of the blocks.

Answer

$\text{R}_1=\text{M}_1\text{g}\cos\theta \ ...(\text{i})$
$\text{R}_2=\text{M}_2\text{g}\cos\theta \ ...(\text{ii})$
$\text{T + M}_1\text{g}\sin\theta-\text{m}_1\text{a}-\mu\text{R}_1=0 \ ...(\text{iii})$
$\text{T}-\text{M}_2-\text{M}_2\text{a}+\mu\text{R}_2=0 \ ...(\text{iv})$
Equn $(iii) \Rightarrow\text{T + M}_1\text{g}\sin\theta-\text{M}_1\text{a}-\mu\text{M}_1\text{g}\cos\theta=0$
Equn $(iv) \Rightarrow\text{T}-\text{M}_2\text{g}\sin\theta+\text{M}_2\text{a}+\mu\text{M}_2\text{g}\cos\theta=0 \ ...(\text{v})$
Equn $(iv) (v) \Rightarrow\text{g}\sin\theta(\text{M}_1+\text{M}_2)-\text{a}(\text{M}_1+\text{M}_2)$
$-\mu\text{g}\cos\theta(\text{M}_1+\text{M}_2)=0$
$\Rightarrow\text{a}(\text{M}_1+\text{M}_2)=\text{g}\sin\theta(\text{M}_1+\text{M}_2)$
$-\mu\text{g}\cos\theta(\text{M}_1+\text{M}_2)$
$\Rightarrow\text{a = g}(\sin\theta-\mu\cos\theta)$
$\therefore$ The blocks system has acceleration $\text{g}(\sin\theta-\mu\cos\theta)$
The force exerted by the rod on one of the blocks is tension.
Tension $\text{T}=-\text{M}_1\text{g}\sin\theta+\text{M}_1\text{a}+\mu\text{M}_1\text{g}\sin\theta$
$\Rightarrow\text{T}=-\text{M}_1\text{g}\sin\theta+\text{M}_1(\text{g}\sin\theta-\mu\text{g}\cos\theta)$
$+\mu\text{M}_1\text{g}\cos\theta$
$\Rightarrow\text{T}=0$

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Question 75 Marks

The angle between the resultant contact force and the normal force exerted by a body on the other is called the angle of friction. Show that, if $\lambda$ be the angle of friction and la the coefficient of static friction, $\lambda\leq\tan^{-1}\mu.$

Answer

$f \rightarrow $ applied force
$F_i \rightarrow $ contact force
$F \rightarrow $ frictional force
$R \rightarrow $ normal reaction
$\mu=\tan\lambda=\frac{\text{F}}{\text{R}}$
When $\text{F}=\mu\text{R, F}$ is the limiting friction $($max friction$)$. When applied force increase, force of friction increase upto limiting friction $(\mu\text{R})$
Before reaching limiting friction

$\text{F}<\mu\text{R}$
$\therefore\tan\lambda=\frac{\text{F}}{\text{R}}\leq\frac{\mu\text{R}}{\text{R}}\Rightarrow\tan\lambda\leq\mu\Rightarrow\lambda\leq\tan^{-1}\mu$

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Question 85 Marks

A body starts slipping down an incline and moves half meter in half second. How long will it take to move the next half meter?

Answer

Suppose, the body is accelerating down with acceleration ‘a’.
From the free body diagram
$\text{R}-\text{mg}\cos\theta=0$
$\Rightarrow\text{R}=\text{mg}\cos\theta \ ...(1)$
$\text{ma + mg}\sin\theta-\mu\text{R}=0$
$\Rightarrow\text{a}=\frac{\text{mg}(\sin\theta-\mu\cos\theta)}{\text{m}}=\text{g}(\sin\theta-\mu\cos\theta)$
For the first half mt. u = 0, s = 0.5m, t = 0.5sec.
So, v = u + at = 0 + (0.5)4 = 2m/s
$\text{S = ut}+\frac{1}{2}\text{at}^2\Rightarrow0.5=0+\frac{1}{2}\text{a}\Big(\frac{0}{5}\Big)^2$
$\Rightarrow\text{a}=4\text{m/s}^2 \ ...(2)$
For the next half metre
$\text{u}=2\text{m/s},\text{ a}=4\text{m/s}^2,\text{ s}=0.5.$
$\Rightarrow0.5=2\text{t}+\Big(\frac{1}{2}\Big)4\text{t}^2\Rightarrow2\text{t}^2+2\text{t}-0.5=0$
$\Rightarrow4\text{t}^2+4\text{t}-1=0$
$\therefore=\frac{-4\pm\sqrt{16+16}}{2\times4}=\frac{1.656}{8}=0.207\text{sec}$
Time taken to cover next half meter is 0.21sec.

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Question 95 Marks

A body of mass 2kg is lying on a rough inclined plane of inclination 30°. Find the magnitude of the force parallel to the incline needed to make the block move:

Up the incline
Down the incline. Coefficient of static friction = 0.2.

Answer

To make the block move up the incline, the force should be equal and opposite to the net force acting down the incline $=\mu\text{R}+2\text{g}\sin30^{\circ}$
$=0.2\times(9.8)\sqrt{3}+219.8\times\Big(\frac{1}{2}\Big)$ [from(1)]
$=3.39+9.8=13\text{N}$
With this minimum force the body move up the incline with a constant velocity as net force on it is zero.

Net force acting down the incline is given by,

$\text{F}=2\text{g}\sin30^{\circ}-\mu\text{R}$
$=2\times9.8\times\Big(\frac{1}{2}\Big)-3.39=6.41\text{N}$
Due to F = 6.41N the body will move down the incline with acceleration.
No external force is required.
$\therefore$ Force required is zero.

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Question 105 Marks

Suppose the entire system of the previous question is kept inside an elevator which is coming down with an acceleration $a < g.$ Repeat parts $(a)$ and $(b).$

Answer

$\text{R}_1+\text{ma}-\text{mg}=0$
$\Rightarrow\text{R}_1=\text{m(g}-\text{a) = mg}-\text{ma} \ ...(\text{i})$
$\text{T}-\mu\text{R}_1=0\Rightarrow\text{T = m(mg}-\text{ma}) \ ...(\text{ii})$
Again, $\text{F}-\text{T}-\mu\text{R}_1=0$
$\Rightarrow\text{F}-\{\mu(\text{mg}-\text{ma})\}-\text{u}(\text{mg}-\text{ma})=0$
$\Rightarrow\text{F}-\mu\text{mg}+\mu\text{ma}-\mu\text{mg}+\mu\text{ma}=0$
$\Rightarrow\text{F}=2\mu\text{mg}-2\mu\text{ma}$
$\Rightarrow\text{F}=2\mu\text{m(g}-\text{a})$

Acceleration of the block be $a_1$

$\text{R}_1=\text{mg}-\text{ma} \ ...(\text{i})$
$2\text{F}-\text{T}-\mu\text{R}_1-\text{ma}_1=0$
$\Rightarrow2\text{F}-\text{t}-\mu\text{mg}+\mu\text{a}-\text{ma}_1=0 \ ...(\text{ii})$

$\text{T}-\mu\text{R}_1-\text{Ma}_1=0$
$\Rightarrow\text{T}=\mu\text{R}_1+\text{Ma}_1$
$\Rightarrow\text{T}=\mu(\text{mg}-\text{ma})+\text{Ma}_1$
$\Rightarrow\text{T}=\mu\text{mg}-\mu\text{ma + Ma}_1$
Subtracting values of $F\ T, $ we get
$2(2\mu\text{m(g}-\text{a}))-2(\mu\text{mg}-\mu\text{ma}+\text{Ma}_1)$
$-\mu\text{mg}+\mu\text{ma}-\mu\text{a}_1=0$
$\Rightarrow4\mu\text{mg}-4\mu\text{ma}-2\mu\text{mg}+2\mu\text{ma = ma}_1+\text{Ma}_1$
$\Rightarrow\text{a}_1=\frac{2\mu\text{m(g}-\text{a})}{\text{M + m}}$
Both blocks move with this acceleration but in opposite directions.

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Question 115 Marks

$A 2\ kg$ block is placed over a $4\ kg$ block and both are placed on a smooth horizontal surface. The coefficient of friction between the blocks is $0.20.$ Find the acceleration of the two blocks if a horizontal force of $12N$ is applied to:

The upper block.
The lower block. Take $g = 10\ m/s^2.$

Answer

$R_1 = 2g = 20$
$\text{Ma}-\mu \text{R}_1=0$
$\Rightarrow 2a = 0.2(20) = 4$
$\Rightarrow a = 2m/s^2$
$4a + 0.2 \times 2 \times 10 - 12 = 0$
$\Rightarrow 4a + 4 = 12$
$\Rightarrow 4a = 8$
$\Rightarrow a = 2m/s^2$

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Question 125 Marks

Consider the situation shown in figure. Calculate:

The acceleration of the $1.0\ kg$ blocks.
The tension in the string connecting the $1.0\ kg$ blocks.
The tension in the string attached to $0.50\ kg.$

Answer

From the free body diagram
$\text{T}+0.5\text{a}-0.5\text{g}=0 \ ...(1)$
$\mu\text{R}+1\text{a}+\text{T}_1-\text{T}=0 \ ...(2)$
$\mu\text{R}+1\text{a}-\text{T}_1=0$
$\mu\text{R}+1\text{a}=\text{T}_1 \ ...(3)$
From $(2) (3)$
$\Rightarrow\mu\text{R + a = T}-\text{T}_1$
$\therefore\text{T}-\text{T}_1=\text{T}_1$
$\Rightarrow\text{T}=2\text{T}_1$
Equation $(2)$ becomes
$\mu\text{R + a + T}_1-2\text{T}_1=0$
$\Rightarrow\mu\text{R + a}-\text{T}_1=0$
$\Rightarrow\text{T}_1=\mu\text{R + a}=0.2\text{g + a} \ ...(4)$
Equation $(1)$ becomes
$2\text{T}_1+\frac{0}{5\text{a}}-0.5\text{g}=0$
$\Rightarrow\text{T}_1=\frac{0.5\text{g}-0.5\text{a}}{2}=0.25\text{g}-0.25\text{a} \ ...(5)$
From $ (4)\ (5) 0.2\text{g + a}=0.25\text{g}-0.25\text{a}$
$\Rightarrow\text{a}=\frac{0.05}{1.25}\times10=0.04/10=0.4\text{m/s}^2$

Accln of $1\ kg$ blocks each is $0.4m/s^2$
Tension $T_1 = 0.2g + a + 0.4 = 2.4N$
$T = 0.5g - 0.5a = 0.5 \times 10 - 0.5 \times 0.4 = 4.8N$

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Question 135 Marks

A block of mass m slips on a rough horizontal table under the action of a horizontal force applied to it. The coefficient of friction between the block and the table is $\mu.$ The table does not move on the floor. Find the total frictional force applied by the floor on the legs of the table. Do you need the friction coefficient between the table and the floor or the mass of the table?

Answer

Because the block slips on the table, maximum frictional force acts on it.
From the free body diagram
$\text{R = mg}$
$\therefore\text{F}-\mu\text{R}=0\Rightarrow\text{F}=\mu\text{R}=\mu\text{mg}$
But the table is at rest. So, frictional force at the legs of the table is not $\mu\text{R}_1.$ Let be f, so form the free body diagram.
$\text{f}_{\text{o}}-\mu\text{R}=0\Rightarrow\text{f}_{\text{o}}=\mu\text{R}=\mu\text{mg}.$
Total frictional force on table by floor is $\mu\text{mg}.$

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Question 145 Marks

A block slides down an inclined surface of inclination 30° with the horizontal. Starting from rest it covers 8m in the first two seconds. Find the coefficient of kinetic friction between the two.

Answer

From the free body diagram,
$\text{R}-\text{mg}\cos\theta=0\Rightarrow\text{R = mg}\cos\theta \ ...(1)$
For the block
U = 0, s = 8m, t = 2sec.
$\therefore\text{s = ut}+\frac{1}{2}\text{at}^2\Rightarrow8=0+\frac{1}{2}\text{a}2^2\Rightarrow\text{a}=4\text{m/s}^2$
Again, $\mu\text{R}+\text{ma}-\text{mg}\sin\theta=0$
$\Rightarrow\mu\text{ mg}\cos\theta+\text{ma}-\text{mg}\sin\theta=0$ [from(1)]
$\Rightarrow\text{m}(\mu\text{g}\cos\theta+\text{a}-\text{g}\sin\theta)=0$
$\Rightarrow\mu\times10\times\cos30^{\circ}=\text{g}\sin30^{\circ}-\text{a}$
$\Rightarrow\mu\times10\times\sqrt{\Big(\frac{3}{3}\Big)}=10\times\Big(\frac{1}{2}\Big)-4$
$\Rightarrow\Big(\frac{5}{\sqrt{3}}\Big)\mu=1\Rightarrow\mu=\frac{1}{\big(\frac{5}{\sqrt{3}}\big)}=0.11$
$\therefore$ Co-efficient of kinetic friction between the two is 0.11.

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Question 155 Marks

If the tension in the string in figure is $16N$ and the acceleration of each block is $0.5\ m/s^2,$ find the friction coefficients at the two contacts with the blocks.

Answer

From the free body diagram
$\mu_1\text{R}+1-16=0$
$\Rightarrow\mu_1(2\text{g})+(-15)=0$
$\Rightarrow\mu_1=\frac{15}{20}=0.75$
$\mu_2\text{R}_1+4\times0.5+16-4\text{g}\sin30^{\circ}=0$
$\Rightarrow\mu_2\Big(\frac{20}{\sqrt{3}}\Big)+2+16-20=0$
$\Rightarrow\mu_2=\frac{2}{20\sqrt{3}}=\frac{1}{17.32}=0.057\approx0.06$
$\therefore$ Co-efficient of friction $\mu_1=0.75$ and $\mu_2=0.06$

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Question 165 Marks

Repeat part (a) of problem 6 if the push is applied horizontally and not parallel to the incline.

Answer

From the free body diagram
$\text{g}=10\text{m/s}^2,\text{m}=2\text{kg},\theta=30^{\circ},\mu=0.2$
$\text{R}-\text{mg}\cos\theta-\text{F}\sin\theta=0$
$\Rightarrow\text{R = mg}\cos\theta+\text{F}\sin\theta \ ...(1)$
And $\text{mg}\sin\theta+\mu\text{R}-\text{F}\cos\theta=0$
$\Rightarrow\text{mg}\sin\theta+\mu(\text{mg}\cos\theta+\text{F}\sin\theta)-\text{F}\cos\theta=0$
$\Rightarrow\text{mg}\sin\theta+\mu\text{ mg}\cos\theta+\mu\text{F}\sin\theta-\text{F}\cos\theta=0$
$\Rightarrow\text{F}=\frac{(\text{mg}\sin\theta+\mu\text{mg}\cos\theta)}{(\mu\sin\theta-\cos\theta)}$
$\Rightarrow\text{F}=\frac{2\times10\times\big(\frac{1}{2}\big)+0.2\times2\times10\times\big(\frac{\sqrt{3}}{2}\big)}{0.2\times\big(\frac{1}{2}\big)-\big(\frac{\sqrt{3}}{2}\big)}$
$=\frac{13.464}{0.76}=17.7\text{N}\approx17.5\text{N}$

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Question 175 Marks

The friction coefficient between the board and the floor shown in figure is $\mu.$ Find the maximum force that the man can exert on the rope so that the board does not slip on the floor.

Answer

Let, the max force exerted by the man is T.
From the free body diagram
$\text{R + T}-\text{Mg}=0$
$\Rightarrow\text{R = Mg}-\text{T} \ ...(\text{i})$
$\text{R}_1-\text{R}-\text{mg}=0$
$\Rightarrow\text{R}_1=\text{R}+\text{mg} \ ...(\text{ii})$
And $\text{T}-\mu\text{R}_1=0$
$\Rightarrow\text{T}-\mu(\text{R + mg})=0$ [From equn. (ii)]
$\Rightarrow\text{T}-\mu\text{R}-\mu\text{mg}=0$
$\Rightarrow\text{T}-\mu(\text{Mg + T})-\mu\text{mg}=0$ [from (i)]
$\Rightarrow\text{T}(1+\mu)=\mu\text{Mg}+\mu\text{mg}$
$\Rightarrow\text{T}=\frac{\mu(\text{M + m)g}}{1+\mu}$
Maximum force exerted by man is $\frac{\mu(\text{M + m)g}}{1+\mu}$

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Question 185 Marks

Find the acceleration of the block of mass $M$ in the situation of figure. The coefficient of friction between the two blocks is $\mu_1$ and that between the bigger block and the ground is $\mu_2.$

Answer

Let the acceleration of block $M$ is $'a\ '$ towards right. So, the block $'m\ '$ must go down with an acceleration $'2a\ '$.

As the block $'m\ '$ is in contact with the block $'M\ '$, it will also have acceleration $'a\ '$ towards right.
So, it will experience two inertia forces as shown in the free body diagram $-1$.
From free body diagram $-1\text{R}_1-\text{ma}=0$
$\Rightarrow\text{R}_1=\text{ma} \ ...(\text{i})$
Again, $2\text{ma + T}-\text{mg}+\mu_1\text{R}_1=0$
$\Rightarrow\text{T = mg}-(2-\mu_1)\text{ma} \ ...(\text{ii})$

From free body diagram $-2\text{T}+\mu_1\text{R}_1+\text{mg}-\text{R}_2=0$
$\Rightarrow\text{R}_2=\text{T}+\mu_1\text{ma + Mg} \ [$Putting the value of $R_1 $ from $(i)]$
$=(\text{mg}-2\text{ma}-\mu_1\text{ma})+\mu_1\text{ma + Mg} \ [$Putting the value of T from $(ii)]$
$\therefore\text{R}_2=\text{Mg + mg}-2\text{ma} \ ...(\text{iii})$
Again, form the free body diagram-2$\text{T + T}-\text{R}-\text{Ma}-\mu_2\text{R}_2=0$
$\Rightarrow2\text{T}-\text{MA}-\text{mA}-\mu_2(\text{Mg + mg}-2\text{ma})=0 \ [$Putting the values of $R_1$ and $R_2$ from $(i)$ and $(iii)]$
$\Rightarrow2\text{T}=\text{(M + m})+\mu_2(\text{Mg + mg}-2\text{ma}) \ ...(\text{iv})$
From equation $(ii)$ and $(iv)\ 2\text{T = 2mg}-2(2+\mu_1)\text{mg}$
$= \text{(M + m)a}+\mu_2(\text{Mg + mg}-2\text{ma})$
$\Rightarrow2\text{mg}-\mu_2(\text{M + m)g}$
$= \text{ a}\text{(M + m}-2\mu_2\text{m}+4\text{m}+2\mu_1\text{m})$
$\Rightarrow\text{a}=\frac{[2\text{m}-\mu_2(\text{M + m})]\text{g}}{\text{M + m}[5+2(\mu_1-\mu_2)]}$

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Question 195 Marks

The friction coefficient between the two blocks shown in figure is $\mu$ but the floor is smooth.

What maximum horizontal force $F$ can be applied without disturbing the equilibrium of the system?
Suppose the horizontal force applied is double of that found in part $(a)$. Find the accelerations of the two masses.

Answer

Both upper block lower block will have acceleration $2m/s^2$

$\text{R}_1=\text{mg} \ ...(\text{i})$
$\text{F}-\mu\text{R}_1-\text{T}=0\Rightarrow\text{F}-\mu\text{mg}-\text{T}=0 \ ...(\text{ii})$
$\therefore\text{F}=\mu\text{mg}+\mu\text{mg}=2\mu\text{mg}$ [putting $\text{T}=\mu\text{mg}$]
$\text{T}-\mu\text{R}_1=0$
$\Rightarrow\text{T}=\mu\text{mg}$

$2\text{F}-\text{T}-\mu\text{mg}-\text{ma}=0 \ ...(\text{i})$
$\text{T}-\text{Ma}-\mu\text{mg}=0$ [$\therefore R_1 = mg]$
$\Rightarrow\text{T = Ma}+\mu\text{mg}$
Putting value of T in (i)$2\text{f}-\text{Ma}-\mu\text{mg}-\mu\text{mg}-\text{ma}=0$
$\Rightarrow2(2\mu\text{mg})-2\mu\text{mg = a(M + m)}$ [Putting $\text{F}=2\mu\text{mg}$]
$\Rightarrow4\mu\text{mg}-2\mu\text{mg = a(M + m)}$
$\Rightarrow\text{a}=\frac{2\mu\text{mg}}{\text{M + m}}$
Both blocks move with this acceleration $'a\ '$ in opposite direction.

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Question 205 Marks

Figure shows two blocks in contact sliding down an inclined surface of inclination $30^\circ .$ The friction coefficient between the block of mass $2.0\ kg$ and the incline is $\mu_1,$ and that between the block of mass $4.0\ kg$ and the incline is $\mu_2.$ Calculate the acceleration of the $2.0\ kg$ block if:

$\mu_1=0.20$ and $\mu_2=0.30$
$\mu_1=0.30$ and $\mu_2=0.20$ Take $\text{g}=10\text{m/s}^2.$

Answer

From the free body diagram

$\text{R = 4g}\cos30^{\circ}=4\times10\times\frac{\sqrt{3}}{2}=20\sqrt{3} \ ...(\text{i})$
$\mu_2\text{R + 4a}-\text{P}-4\text{g}\sin30^{\circ}=0$
$\Rightarrow0.3(40)\cos30^{\circ}+4\text{a}-\text{P}-40\sin20^{\circ}=0 \ ...(\text{ii})$
$\text{P + 2a}+\mu_1\text{R}_1-2\text{g}\sin30^{\circ}=0 \ ...(\text{iii})$
$\text{R}_1=2\text{g}\cos30^{\circ}=2\times10\times\frac{\sqrt{3}}{2}=10\sqrt{3} \ ...(\text{iv})$
Equn.$(ii) 6\sqrt{3}+4\text{a}-\text{P}-20=0$
Equn $(iv) \text{P + 2a}+2\sqrt{3}-10=0$
From Equn $(ii)\ (iv) 6\sqrt{3}+6\text{a}-30+2\sqrt{3}=0$
$\Rightarrow6\text{a}=30-8\sqrt{3}=30-13.85=16.15$
$\Rightarrow\text{a}=\frac{16.15}{6}=2.69=2.7\text{m/s}^2$
can be solved. In this case, the $4\ kg$ block will travel with more acceleration because, coefficient of friction is less than that of $2\ kg.$ So, they will move separately. Drawing the free body diagram of $2\ kg$ mass only, it can be found that, $a = 2.4m/s^2.$

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Question 215 Marks

The friction coefficient between a road and the tyre of a vehicle is $\frac{4}{3}.$ Find the maximum incline the road may have so that once hard brakes are applied and the wheel starts skidding, the vehicle going down at a speed of 36km/hr is stopped within 5m.

Answer

$\text{s = 5m}, \ \mu=\frac{4}{3},\text{g}=10\text{m/s}^2$
$\text{u}=36\text{km/h}=10\text{m/s},\text{ v}=0,$
$\text{a}=\frac{\text{v}^2-\text{u}^2}{2\text{s}}=\frac{0-10^2}{2\times5}=-10\text{m/s}^2$
From the freebody diagrams,
$\text{R}-\text{mg}\cos\theta=0;\text{g}=10\text{m/s}^2$
$\Rightarrow\text{R = mg}\cos\theta \ ...(\text{i});\mu=\frac{4}{3}.$
Again, $\text{ma + mg}\sin\theta-\mu\text{R}=0$
$\Rightarrow\text{ma + mg}\sin\theta-\mu\text{mg}\cos\theta=0$
$\Rightarrow\text{a + g}\sin\theta-\text{mg}\cos\theta=0$
$\Rightarrow10+10\sin\theta-\Big(\frac{4}{3}\Big)\times10\cos\theta=0$
$\Rightarrow30+30\sin\theta-40\cos\theta=0$
$\Rightarrow3+3\sin\theta-4\cos\theta=0$
$\Rightarrow4\cos\theta-3\sin\theta=3$
$\Rightarrow4\sqrt{1-\sin^2\theta}=3+3\sin\theta$
$\Rightarrow16(1-\sin^2\theta)=9+9\sin^2\theta+18\sin\theta$
$\sin\theta=\frac{-18\pm\sqrt{18^2-4(25)(-7)}}{2\times25}\\=\frac{-18\pm32}{50}=\frac{14}{50}=0.28$ [Taking +ve sign only]
$\Rightarrow\theta=\sin^{-1}(0.28)=16^{\circ}$
Maximum incline is $\theta=16^{\circ}$

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Question 225 Marks

Figure shows a small block of mass m kept at the left end of a larger block of mass $M$ and length $l.$ The system can slide on a horizontal road. The system is started towards right with an initial velocity $v.$ The friction coefficient between the road and the bigger block is $g$ and that between the block is $\frac{\mu}{2}.$ Find the time elapsed before the smaller blocks separates from the bigger block.

Answer

Let $a_1$ and $a_2$ be the accelerations of ma and $M$ respectively.
Here, $a_1 > a_2$ so that $m$ moves on $M$
Suppose, after time $‘t\ ’\ m$ separate from $M.$
In this time, $m$ covers $\text{vt}+\frac{1}{2}\text{ a}_1\text{ t}^2$ and $\text{S}_{\text{M}}=\text{vt}+\frac{1}{2}\text{ a}_2\text{ t}^2$
For $‘m\ ’$ to $m$ to $‘m\ ’$ separate from $M. \text{vt}+\frac{1}{2}\text{ a}_1\text{ t}^2=\text{vt}+\frac{1}{2}\text{ a}_2\text{ t}^2+\ell \ ...(1)$

Again from free body diagram
$\text{Ma}_1+\frac{\mu}{2}\text{R}=0$
$\Rightarrow\text{ma}_1=-\Big(\frac{\mu}{2}\Big)\text{mg}$
$=-\Big(\frac{\mu}{2}\Big)\text{m}\times10\Rightarrow\text{a}_1=-5\mu$
Again,
$\text{Ma}_2+\mu(\text{M + m)g}-\Big(\frac{\mu}{2}\Big)\text{mg}=0$
$\Rightarrow2\text{Ma}_2+2\mu(\text{M + m)g}-\mu\text{mg}=0$
$\Rightarrow2\text{Ma}_2=\mu\text{mg}-2\mu\text{Mg}-2\mu\text{mg}$
$\Rightarrow\text{a}_2\frac{-\mu\text{mg}-2\mu\text{Mg}}{2\text{M}}$
Putting values of $a_1 a_2$ in equation $(1)$ we can find that
$\text{T}=\sqrt{\Big(\frac{4\text{ml}}{(\text{M + m})\mu\text{g}}\Big)}$

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5 Marks Questions - Physics STD 12 Science Questions - Vidyadip