[3 Mark Question Answer]

Question 13 Marks

How is weight affected in the following cases, when initially the body is weighed in Delhi with a spring balance?
1. Body is taken to Moscow.
2. Body is taken to Ceylon.
3. Body is taken to sea level.
4. Body is taken to a high mountain.
5. Body is taken deep inside a mine.

Answer

1. When body is taken to Moscow, then weight of the body weighs slightly more than Delhi.
2. When body is taken to Ceylon, then weight of the body is slightly more than at Delhi.
3. When body is taken to sea level, then its weight remains same.
4. As the value of $g$ increases with increase in height of body from surface of the earth, so when a body is taken .to a high mountain, its weight decreases.
5. As the value of $g$ decreases with depth, so when body is taken deep inside a mine, its weight decreases.

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Question 23 Marks

A man weighs 800 N the at the equator. How does the weight of man change at the following places?
(a) At poles
(b) 100 km up in space
(c) 10 km down in a mine.

Answer

Weight of the man at equator $=800 N$
(a) At poles: As value of g increases at poles as compared to that at equator, so weight of the body at poles will be more than 800 N .
(b) As the value of g decreases with increase in height, so weight of the man 100 km up in the space is less than 800 N .
(c) As value of g decreases with increase in depth, so weight of the man 10 km down in a mine is less than 800 N .

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Question 33 Marks

A man sits in a machine which generates acceleration five times more than acceleration due to gravity. If the mass of man is 80 kg , what is his weight? Take $g =10 ms^{-2}$.

Answer

$\begin{array}{l}\text { Mass of the man }= m =80 kg \\ \text { Acceleration due to gravity }= g =10 ms^{-2} \\ \text { Acceleration generated by the machine } \\ = a =5 g=5 \times 10=50 ms^2 \\ \text { Case-I: } \\ \text { When machine accelerates downward } \\ \text { Effective weight of the man = } mg - ma = m ( g - a ) \\ =80(50-10)=3200 N \\ \text { Case-II: When the machine accelerates upward } \\ \text { Effective weight of man = } mg + ma = m ( g + a ) \\ =80(10+50) \\ =80 \times 60=4800 N\end{array}$

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Question 43 Marks

State four differences between mass and weight.

Answer

Mass
1. It is the quantity of matter contained in a body.
2. It is constant quantity.
3. It is a scalar quantity.
4. Its unit is kg.
5. Beam balance is used to measure it.
6. At no place its value is zero.
Weight:
1. It is the force with which a body is attracted towards the center of the earth.
2. $W = mxg$, hence its value changes from place to place according to the value of 'g'
3. It is a vector quantity.
4. Its unit is 'newton'.
5. Spring balance is used to measure it.
6. At the center of earth its value is zero.

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Question 53 Marks

(a) What do you understand by the term weight?
(b) State two important characteristics of weight
(c) State the units of weight in CGS and Sl system.
(d) Name the device used for measuring weight

Answer

(a) Weight: Weight of a body is defined as the force with which the earth attracts it.
(b) Characteristics of weight:
1. It depends upon the position and surroundings of body.
2. It is a vector quantity.
3. If changes from place to place on the surface of earth due to the change in the value of acceleration due to gravity (g).
(c) In CGS system, unit of weight is dyne. In Sl system, unit of weight is newton (N).
(d) Spring balance is used to measure the weight of the body.

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Question 63 Marks

(a) What do you understand by the term mass?
(b) State two important characteristics of mass.
(c) State units of mass in CGS and SI systems.
(d) Name the device used for measuring mass.

Answer

(a) Mass: The quantity of matter contained in a body is known as its mass.
(b) Characteristics of mass:
1. It is independent of the position and surrounding of a body.
2. It is a scalar quantity.
3. It remains constant at all places, provided the velocity of the body is not too high.
(c) CGS unit of mass is gram ( g ).
Sl unit of mass is kilogram (kg).
(d) Physical balance is used to measure the mass of a body.

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Question 73 Marks

State Newton's law of gravitation.

Answer

This law states that the force of attraction acting between two particles is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.

i.e. $F \propto m_1 m_2$
And $F \propto \frac{1}{r^2}$
Combining (i) and (ii), we have
$
\begin{array}{l}
F \propto \frac{m_1 m_2}{r^2} \\
\text { or } F=\frac{G m_1 m_2}{r^2}
\end{array}
$
Where G is the constant of proportionality and is known as the universal gravitational constant. Its value at all places in this universe is $6.67 \times 10^{-11} Nm ^2 kg^{-2}$.

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Question 83 Marks

Define absolute units of force in CGS as well as SI system.

Answer

Absolute unit of force in CGS system is dyne and in SI system is Newton (N).
One dyne: When the body of mass 1 gram moves with an acceleration of $1 cms ^{-2}$, then the force acting on the body is called one dyne.
1 dyne $=1 g cms ^{-2}$
One Newton: When a body of mass 1 kg moves with an acceleration of $1 ms^{-2}$, then force acting on the body is said to be one newton.
OR
That force is said to be one newton, which producers an acceleration of $1 ms^{-2}$ in a body of mass 1 kg .
$1 N=1 kg ms ^{-2}$

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Question 93 Marks

A force of 600 dynes acts on a glass ball of mass 200 g for 12 s . If initially the ball is at
rest,
find
(1) Final velocity
(2) Distance covered.

Answer

$
\begin{array}{l}
F=600 \text { dyne }=600 g cms^{-2} \\
m=200 g
\end{array}
$
$
\text { Acceleration }=a=\frac{F}{m}=\frac{600 g cms^{-2}}{200 g}
$
$a -3 cms ^{-2}$
Also time $= t =12 s$
Initial velocity $= u =0$
(1) Final velocity $=v=$ ?
$
\begin{array}{l}
v=u+a t \\
v=0+3(12) \\
v=36 cms^{-1}
\end{array}
$
(2) Distance covered $= S =$ ?
$
\begin{array}{l}
v^2-w^2=2 a s \\
(36)^{2-(0)^2}=2(3) s \\
S=\frac{ 3 6 \times 3 6 }{ 6 }=36 \times 6=216 cm
\end{array}
$

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Question 103 Marks

A cricket player holds a cricket ball of mass 100 g by moving his hands backward by 0.75 m . If the initial velocity of the ball is $108 kmh ^{-1}$, find the retarding force applied by the player.

Answer

Mass of ball $- m =100 g=0.1 kg$
Distance covered $= S =0.75 m$
Initial velocity of ball $= u =108 kmh ^{-1}$
$
u=108 \times \frac{5}{18} ms^{-1}=30 ms^{-1}
$
Final velocity of car $= v =0$
Retarding force $= F =$ ?
$
\begin{array}{l}
v^2-w^2=2 a S \\
(0)^2-(30)^2=2 a(0.75) \\
1.5 a=-900 \\
\quad a=-\frac{900}{1.5}=-600 ms^{-2}
\end{array}
$
Retarding force $= F = ma$
$
\begin{array}{l}
F=0.1 \times(-600) \\
F=-60 N
\end{array}
$

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Question 113 Marks

A car of mass 800 kg , moving at $54 kmh ^{-1}$ is brought to rest over a distance of 15 m . Find the retarding force developed by the brakes of the car.

Answer

Mass of car $=m=800 kg$
Initial velocity of car $= u -54 kmh ^{-1}$
$
u=54 \times \frac{5}{18} ms^{-1}=15 ms^{-1}
$
Final velocity of car $=v=0$
Distance covered $= S =15 m$
Retarding force $= F =$ ?
$
\begin{array}{l}
v^2-w^2=2 a S \\
(0)^2-(15)^2=2 a(15) \\
30 a=-225 \\
a =-\frac{ 2 2 5 }{ 3 0 } \\
a=-7.5 ms^{-2} \\
F=ma=(800)(-7.5) \\
F=-6000 N
\end{array}
$

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Question 123 Marks

A golfer hits a ball at rest, such that the contact between the ball and golf stick is for 0.1 s . If the golf ball covers a linear distance of 400 m in 2 s , find the magnitude of force applied. Mass of golf ball is 50 g .

Answer

Distance covered by ball $=400 m$
Time taken $=2 s$
Distance 400
$
\therefore \text { Uniform velocity of the ball }=\frac{\text { Distance }}{\text { Time }}=\frac{400}{2}=200 ms^{-1}
$
Now initial velocity of the ball $= u =0$
Time for which force acts on the ball $=t=0.1 s$
Final velocity of the ball after the force stops acting on it
$
=v=200 ms^{-1}
$
Using, $v = ut + at$
We have, $200=0+a(0.1)$
$
a=\frac{200}{0.1}=2000 ms^{-2}
$
Mass of the ball $= m =50 g=0.05 kg$
Force acting on the ball $= F = ma$
$
\begin{array}{l}
F=0.05 \times 2000 \\
F=100 N
\end{array}
$

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Question 133 Marks

A car initially at rest, picks up a velocity of $72 kmh ^{-1}$ in 20 seconds. If the mass of the car is 1000 kg , find (1) Force developed by its engine
(2) Distance covered by the car.

Answer

Initial velocity $=u=0$
Final velocity $=v=72 kmh ^{-1}$
$v=72 \times \frac{5}{18} ms^{-1}=20 ms^{-1}$
Mass of car $= m =1000 kg$
Time $= t =20 s$
(1)
$
\begin{array}{l}
v=u+a t \\
20=0+a(20)
\end{array}
$
(Acceleration) $a=\frac{20}{20}=1 ms^{-2}$
Now, F = ma
$
F=1000 \times 1=1000 N
$
(2) Distance covered by car $= S =$ ?
$
\begin{aligned}
S & =u t+\frac{1}{2} a t^2 \\
& =0(20)+\frac{1}{2}(1)(20)^2=0+\frac{400}{2} \\
S & =200 m
\end{aligned}
$

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Question 143 Marks

A motorcycle of mass 100 kg is running at $10 ms^{-1}$. If its engine develops an extra linear momentum of 2000 Ns , calculate the new velocity of motorcycle.

Answer

Mass of motor cycle $= m =100 kg$
Velocity of motor cycle $= v _1=10 ms^{-1}$
Momentum of motor cycle $= mv _1=100 \times 10=1000 Ns$
When engine develops an extra linear momentum of2000 Ns
Then total momentum of motor cycle $=1000+2000=3000$ Ns
Let $v_2=$ new velocity of the motor cycle.
Total momentum of motor cycle $= mv _2$
$
3000=100 \times v_2
$
$
v_2=\frac{3000}{100}=30 ms^{-1}
$

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