Physics Test

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Physics Test - 19

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Weekly Quiz Competition

Question 1
1 / -0

A motor car moving with a uniform speed of 20 ms⁻¹ comes to rest on the application of brakes after travelling a distance of 10 m. Its acceleration is

A
20m/sec²

B
−20m/sec²

C
−40m/sec²

D
+2m/sec²

Solution

Using, \(v^{2}=u^{2}+2 a S\)
\(\Rightarrow o=u^{2}+2 a S\)
\(\Rightarrow a=-\frac{u^{2}}{2 S}\)
\(a=-\frac{(20)^{2}}{2 \times 10}\)
\(a=-20 m s^{-2}\)
Question 2
1 / -0

Using the expression 2dsinθ=λ, one calculates the values of d by measuring the corresponding angles θ in the range θ to 90^o. The wavelength λ is exactly known and the error in θ is constant for all values of θ. As θ increases from 0^o,

A
the absolute error in d remains constant

B
the absolute error in d increases

C
the fractional error in d remains constant

D
the fractional error in d decreases

Solution

We have \(2 d \sin \theta=\lambda\)
Let \(2 d=y\) Thus we get, \(y \sin \theta=\lambda\)
\(\Rightarrow \ln y+\ln \sin \theta=\ln \lambda\)
Differentiating both the sides we get \(\Rightarrow \frac{d y}{y}=-\cot \theta\)
\(\Rightarrow\left|\frac{d y}{y}\right|=\cot \theta\)
As \(y=2 d,\) we see that the fractional error in decreases as \(\theta\) increases.
Question 3
1 / -0

The magnitude of the resultant of two equal vectors is equal to the magnitude of either vector. What is the angle between the two vectors?

A
60^∘

B
90^∘

C
120^∘

D
150^∘

Solution

Solution \(R^{2}=A^{2}+B^{2}+2 A B \cos \theta .\) It is given that \(R=A=B\) Putting these values we have \(A^{2}=A^{2}+A^{2}+2 A^{2} \cos \theta\)
or
\[
\cos \theta=-\frac{1}{2} \text { which gives } \theta=120^{\circ}
\]
Hence the correct choice is (c)
Question 4
1 / -0

A rod of mass \(M,\) hinged at \(O\) is kept in equilibrium with a spring of stiffness \(k\) as shown. The

potential energy stored in the spring is:

A
(mg)²/4k

B
(mg)2/2k

C
(mg)2/8k

D
(mg)2/k

Solution
Question 5
1 / -0

A motorcycle is going on an overbridge of radius R . The driver maintains a constant speed. As the motorcycle is ascending on the overbridge, the normal force on it

A
Increases

B
Decreases

C
Remains the same

D
Fluctuates

Solution
Question 6
1 / -0

A 10 kg mass is resting on a horizontal surface and horizontal force of 80 N is applied. If μ=0.2, the ratio of acceleration without friction and with friction is (g=10ms⁻²)

A
3/4

B
4/3

C
1/2

D
2

Solution

\(\frac{a_{1}}{a_{2}}=\frac{\frac{F}{m}}{\frac{F-\mu m g}{m}}\)
\(a_{1}=\frac{F}{m}=\frac{80}{10}=8 \mathrm{m} / \mathrm{s}^{2}\)
\(a_{2}=\frac{F-\mu \mathrm{mg}}{m}=\frac{80-0.2 \times 10 \times 10}{10}=6 \mathrm{m} / \mathrm{s}^{2}\)
\(\therefore \frac{a_{1}}{a_{2}}=\frac{8}{6}=\frac{4}{3}\)
Question 7
1 / -0

Modulus of elasticity for a perfectly elastic body is

A
Zero

B
Infinity

C
1

D
Can have any value

Solution

For a perfect elastic body the stress strain graph is always linear i.e stress is always proportional to the strain through a constant i.e., the young's modulus of elasticity which is proportional to the slope of the linear property of the stress strain graph. if a body is ideally plastic it means to have infinite linear line from the center of the stress strain graph. A linear line like this could have any slope=tan θ.this linear line could have any angle meaning any slope can values between 1 and 0 implying that youngs modulus could be any value.
Question 8
1 / -0

A stone dropped from the top of the tower touches the ground in 4 sec. The height of the tower is about

A
80 m

B
40 m

C
20 m

D
160 m

Solution

We have,
\[
h=\frac{1}{2} g t^{2}=\frac{1}{2} \times 10 \times(4)^{2}=80 m
\]
Question 9
1 / -0

A body of mass 10 kg is slipping on a rough horizontal plane and moves with a deceleration of 4.0 m/s². What is the magnitude of frictional force?

A
40N

B
20N

C
100N

D
Data insufficient require coefficient of kinetic friction

Solution

As friction can be the only external force acting of the body in horizontal direction
Question 10
1 / -0

A ball moving with a speed of 2.2 m/sec strikes an identical stationary ball. After collision the first ball moves at 1.1 m/sec at 60⁰ with the original line of motion. The magnitude and direction of the ball after collision is

A
5m/sec, 90^o

B
2m/sec, 60^o

C
√39(1.1) m/sec, 30^o

D
10m/sec, 60o

Solution

Conserving momentum in \(\mathrm{X}\) - direction \(m \times 2.2=1.1 \times m \times \cos 60+v \cos \theta \times m\)
\(2.2=\frac{1.1}{2}+v \cos \theta\)
\(\frac{3.3}{2}=v \cos \theta\)
Similarly in \(Y\) - direction
\(v \sin \theta=1.1 \sin 60\)
\(v \sin \theta=\frac{1.1 \sqrt{3}}{2}\)
from (1) \(8 x(2)\)
\[
v=\sqrt{3.9(1.1)} \mathrm{m} / \mathrm{s}
\]
\(\boldsymbol{\theta}=30^{0}\)

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Re-Attempt Weekly Quiz Competition

Physics Test - 19

Result

Physics Test - 19

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SHARING IS CARING

Question 1

20m/sec2

−20m/sec2

−40m/sec2

+2m/sec2

Solution

Question 2

the absolute error in d remains constant

the absolute error in d increases

the fractional error in d remains constant

the fractional error in d decreases

Solution

Question 3

60∘

90∘

120∘

150∘

Solution

Question 4

(mg)2/4k

(mg)2/2k

(mg)2/8k

(mg)2/k

Solution

Question 5

Increases

Decreases

Remains the same

Fluctuates

Solution

Question 6

3/4

4/3

1/2

2

Solution

Question 7

Zero

Infinity

1

Can have any value

Solution

Question 8

80 m

40 m

20 m

160 m

Solution

Question 9

40N

20N

100N

Data insufficient require coefficient of kinetic friction

Solution

Question 10

5m/sec, 90o

2m/sec, 60o

√39(1.1) m/sec, 30o

10m/sec, 60o

Solution

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20m/sec²

−20m/sec²

−40m/sec²

+2m/sec²

60^∘

90^∘

120^∘

150^∘

(mg)²/4k

5m/sec, 90^o

2m/sec, 60^o

√39(1.1) m/sec, 30^o