Physics Test

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Physics Test - 43

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Weekly Quiz Competition

Question 1
1 / -0

A force of 10N acts on a body of mass 2kg for 3s, initially at rest. Calculate: (i) the velocity acquired by the body, and (ii) change in momentum of the body.

A
15 m s⁻¹30 kg m s⁻¹

B
15 m s⁻¹ 20 kg m s⁻¹

C
5 m s⁻¹ 30 kg m s⁻¹

D
25 m s⁻¹ 30 kg m s⁻¹

Solution

\(F=10 N, m=2 k g, \Delta t=3 s, u=0\)
(i) \(F=m a\) Putting the values, \(10 N=2 k g \times(a)\) Therefore, acceleration, \(a=5 m / s^{2}\) Acceleration, \(a=\frac{\Delta v}{\Delta t}\)
Change in velocity, \(\Delta v=a \times \Delta t=5 m / s^{2} \times 3 s=15 m / s\)
\(v-u=15 m / s\)
Therefore, velocity acquired by the body, \(v=15 \mathrm{m} / \mathrm{s}\)
(ii) Force can be also written as, \(F=\frac{\Delta p}{\Delta t}\) Where, \(p\) is momentum. Therefore, change in momentum, \(\Delta p=F \times \Delta t=10 N \times 3 s=30 \mathrm{kgm} / \mathrm{s}\)
Hence, the correct option is (A)
Question 2
1 / -0

One Newton is the force, which produces an acceleration of :

A
1ms⁻²on a body of mass 1gm

B
1cms⁻² on a body of mass 1kg

C
1cms⁻² on a body of mass 1gm

D
1ms⁻² on a body of mass 1kg

Solution

Newton is unit of force in SI system: 1 Newton is the force needed to accelerate \(1 \mathrm{kg}\) of mass at the rate of \(1 \mathrm{m} / \mathrm{s}^{2}\) We know that Force\(=\)mass\(\times\)acceleration
where S.I unit of mass is \(k g\) and S.I. unit of acceleration is \(m / s^{2}\) \(\therefore 1 N=1 k g \times 1 m / s^{2}\)
Hence, the correct option is (D)
Question 3
1 / -0

If a constant force acts on a body initially at rest, the distance moved by the body in time t is proportional to:

A
1/t

B
t

C
t²

D
t³

Solution

\(F=m a=\) constant
\(\Longrightarrow a=\) constant
Given \(u=0\) For uniform acceleration, \(s=u t+\frac{1}{2} a t^{2}\)
\(s=\frac{1}{2} a t^{2}\)
\(s \propto t^{2}\)
Hence, the correct option is (C)
Question 4
1 / -0

A bullet penetrates a distance d of a plank. If the initial momentum of the bullet is doubled, the distance penetrated will be

A
4d

B
d

C
2d

D
d/2

Solution

Answer: \(-\mathrm{A}\)
Initial momentum - \(\boldsymbol{p}_{1}\) Final momentum - \(\boldsymbol{p}_{2}\) Distance penetrated - \(\boldsymbol{x}\)
\(p_{1}=m v\)
\(0^{2}=v^{2}-2 a d\)
\(v=\sqrt{2 a d}\)
\(p_{2}=2 p_{1}=2 m \sqrt{2 a d}=m v_{2}\)
\(v_{2}=2 \sqrt{2 a d}\)
\(0^{2}=(2 \sqrt{2 a d})^{2}-2 a x\)
\(x=4 d\)
Hence, the correct option is (A)
Question 5
1 / -0

A body of mass 2 kg is at rest. What is the impulse required to impart a velocity of 8 m/s to it in 1 s?

A
40 Ns

B
80 Ns

C
20 Ns

D
16 Ns

Solution

ImpulseP= change in momentum

So, p=2×(8−0)=16 Ns
Hence, the correct option is (D)
Question 6
1 / -0

In the system shown in figure all surfaces are smooth. String is massless and inextensible. Find acceleration a of the system and tension T in the string. ( g=10m/s²)

A
40/3N

B
20N

C
25N

D
50N

Solution

Here, weight of \(2 k g\) is perpendicular to motion (or \(a\) ). Hence, it will not contribute in net pulling force. Only weight of \(4 k g\) block will be included. \(\therefore a=\frac{\text {Net pulling force}}{\text {Total mass}}\)
\(=\frac{(4)(10)}{(4+2)}\)
\(=\frac{20}{3} m / s^{2}\)
For \(T\), consider FBD of \(4 k g\) block. Writing equation of motion. \(40-T=4 a\)
\(\therefore T=40-4 a\)
\(=40-4 \frac{20}{3}\)
\(=\frac{40}{3} N\)
Hence, the correct option is (A)
Question 7
1 / -0

In the given figure, the pulley is massless and frictionless. There is no friction between the body and the floor. The acceleration produced in the body when it is displaced through a certain distance with force \(P\) will be:

A

\(\frac{P}{M}\)

B
\(\frac{P}{4M}\)

C
\(\frac{P}{2M}\)

D
\(\frac{P}{3M}\)

Solution

In this problem, total tension(T) balance by the force (P). so, \(2 T=P, T=\frac{P}{2}\)
If \(a\) is the acceleration of the body of mass \(M\), then:
\(\Rightarrow M a=T\)
\(\Rightarrow a=\frac{T}{M}=\frac{P}{2 M}\)
Hence, the correct option is (C)
Question 8
1 / -0

A person pushing a wall hard is liable to fall back. Which law is related for this ?

A
Newton's first law

B
Newton's second law

C
Newton's third law

D
Newton's gravitational law

Solution

If a person applies a force F on the wall, wall applies the same force on the person in opposite direction. Since wall is very massive as compared to the person, it doesn't move, but the person is likely to fall back because of the reaction force.
Hence, the correct option is (C)
Question 9
1 / -0

Which of the following statement is correct?

A
The area of displacement - time graph gives velocity.

B
The slope of velocity - time graph gives acceleration.

C
The slope of displacement - time graph gives acceleration.

D
The area of velocity - time graph gives average velocity.

Solution

The slope of displacement time graph gives the velocity by equation Velocity \(\vec{v}=\frac{d \vec{x}}{d t}\)
The slope of velocity and time time graph gives acceleration by equation acceleration \(\vec{a}=\frac{d \vec{v}}{d t}\) The area of velocity time graph gives the displacement not the average velocity.
Hence, the correct option is (B)
Question 10
1 / -0

The path of one projectile as seen from another projectile is a :

A
Straight line

B
Parabola

C
Hyperbola

D
Circle

Solution

Acceleration of a projectile is constant (equal to the acceleration due to gravity, g) towards the earth. Hence, acceleration of both projectiles is same and relative acceleration is zero. Since relative acceleration is zero, relative velocity is constant and hence, the path of a projectile seen from another projectile is a straight line.
Hence, the correct option is (A)

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Physics Test - 43

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Physics Test - 43

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Question 1

15 m s−130 kg m s−1

15 m s−1 20 kg m s−1

5 m s−1 30 kg m s−1

25 m s−1 30 kg m s−1

Solution

Question 2

1ms−2 on a body of mass 1gm

1cms−2 on a body of mass 1kg

1cms−2 on a body of mass 1gm

1ms−2 on a body of mass 1kg

Solution

Question 3

1/t

t

t2

t3

Solution

Question 4

4d

d

2d

d/2

Solution

Question 5

40 Ns

80 Ns

20 Ns

16 Ns

Solution

Question 6

40/3N

20N

25N

50N

Solution

Question 7

\(\frac{P}{M}\)

\(\frac{P}{4M}\)

\(\frac{P}{2M}\)

\(\frac{P}{3M}\)

Solution

Question 8

Newton's first law

Newton's second law

Newton's third law

Newton's gravitational law

Solution

Question 9

The area of displacement - time graph gives velocity.

The slope of velocity - time graph gives acceleration.

The slope of displacement - time graph gives acceleration.

The area of velocity - time graph gives average velocity.

Solution

Question 10

Straight line

Parabola

Hyperbola

Circle

Solution

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15 m s⁻¹30 kg m s⁻¹

15 m s⁻¹ 20 kg m s⁻¹

5 m s⁻¹ 30 kg m s⁻¹

25 m s⁻¹ 30 kg m s⁻¹

1ms⁻²on a body of mass 1gm

1cms⁻² on a body of mass 1kg

1cms⁻² on a body of mass 1gm

1ms⁻² on a body of mass 1kg

t²

t³