Physics Test

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Physics Test - 27

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Weekly Quiz Competition

Question 1
1 / -0

Stretching of a rubber band results in ___________.

A
No change in potential energy

B
Zero value of potential energy

C
Increase in potential energy

D
Decrease in potential energy

Solution

When a rubber band is stretched, work is done on it , in course of which some amount of energy in the form of elastic potential energy (strain energy) is input into it. This increases the potential energy of the system. When the rubber band is released, the potential energy is quickly converted to kinetic energy.

Option C.
Hence, the correct option is (C)
Question 2
1 / -0

A wire whose cross-section area is 4mm² is stretched by 0.1mm by a certain weight. How far will a wire of the same material and length stretch, if its cross-sectional area is 8mm²and the same weight is attached?

A
0.5mm

B
1.0mm

C
0.05mm

D
0.06mm

Solution

Young's modulus is given by:
\(Y=\frac{F L}{A \Delta L}\)
It remains the same for both cases as it is a property of the material. \(\frac{F L}{A_{1} \Delta L_{1}}=\frac{F L}{A_{2} \Delta L_{2}}\)
\(\Delta L_{2}=\frac{4}{8} \times 0.1=0.05 \mathrm{mm}\)
Hence, the correct option is (C)
Question 3
1 / -0

One end of a horizontal thick copper wire of length 2L and radius 2R is welded to an end of another horizontal thin copper wire of length L and radius R. When the arrangement is stretched by applying forces at two ends, the ratio of the elongation in the thin wire to that in the thick wire is:

A
0.25

B
0.50

C
2.00

D
4.00

Solution

We have change in length \(\Delta l=\frac{F \cdot l}{Y A}\)
since the two rods are in series \(\left(F_{1}\right)_{\text {rest}}=\left(F_{2}\right)_{\text {rest}}\)
\(\triangle l \propto \frac{l}{R^{2}} \quad\left(\because A=\pi R^{2}\right)\)
\(\therefore \frac{\Delta l_{1}}{\Delta l_{2}}=\frac{l_{1}}{l_{2}} \times \frac{R_{2}^{2}}{R_{1}^{2}}\)
\(\frac{\Delta l_{1}}{\Delta l_{2}}=\frac{2 L}{L} \times \frac{R_{1}^{2}}{4 R_{1}^{2}}\)
\(\frac{\Delta l_{1}}{\Delta l_{2}}=\frac{1}{2}\)
\(\therefore \frac{\Delta l_{2}}{\Delta l_{1}}=\frac{2}{1}=2\)
Hence, the correct option is (C)
Question 4
1 / -0

Let L be the length and d be the diameter of cross-section of a wire. Wires of the same material with different L and d are subjected to the same tension along the length of the wire. In which of the following cases, the extension of wire will be the maximum?

A
L=200 cm,d=0.5 mm

B
L=300 cm,d=1.0 mm

C
L=50 cm,d=0.05 mm

D
L=100 cm,d=0.2 mm

Solution

since material is same, Young's modulus is also same. We have, \(Y \frac{\Delta l}{l}=\frac{F}{A} \Rightarrow \Delta l=\frac{F l}{Y A} \propto \frac{l}{d^{2}}\)
where, \(Y\) is the Young's Modulus, \(F\) is the tension which is same in all cases, \(A\) is the area of cross section of the wire and \(l\) is the length of the wire.
Calculating the value of \(\frac{l}{d^{2}}\) in all cases, we gett
A: \(\frac{l}{d^{2}}=\frac{2000}{0.5 \times 0.5}=8000\)
\(\mathrm{B}: \frac{l}{d^{2}}=\frac{3000}{1.0 \times 1.0}=3000\)
C \(_{t} \frac{l}{d^{2}}=\frac{50}{0.05 \times 0.05}=20000\)
\(\mathrm{D}_{4} \frac{l}{d^{2}}=\frac{100}{0.2 \times 0.2}=2500\)
So maximum elongation will be in case C.
Hence, the correct option is (C)
Question 5
1 / -0

A wire can be broken by applying load of 200N. The force required to break another wire of the same length and same material, but double in diameter is

A
200 N

B
400 N

C
600 N

D
800 N

Solution

Since the wires have same material, their modulus of elasticity must be same.

Thus, the ratio of longitudinal stress to longitudinal strain (which is Young's Modulus of Elasticity) must be same. Now, before breaking, the strains in both wires must be the same too.

Therefore, stress == Young's modulus × strain, must be same for both wires.

The second wire has diameter double that of the first wire, so area of cross section of second wire is 4 times as large as that of first wire. Therefore, to develop the same stress, the force applied on second wire must be 4 times as large as the force applied on the first wire (since stress = force/cross section area).

Thus, force that needs to be applied to second wire to break it is 4×200=800N
Hence, the correct option is (D)
Question 6
1 / -0

How much force is required to produce an increase of 0.2% in the length of a brass wire of diameter 0.6 mm? (Young's modulus for brass =0.9×10¹¹N/m²)

A
Nearly 17 N

B
Nearly 34 N

C
Nearly 51 N

D
Nearly 68 N

Solution

From Young's modulus relation, \(Y=\frac{F / A}{\Delta L / L}\)
\(\Longrightarrow F=Y A\left(\frac{\Delta L}{L}\right)\)
\(=0.9 \times 10^{11} \times \pi(0.0003)^{2} \times 0.002=50.8 N \approx 51 N\)
Hence, the correct option is (C)
Question 7
1 / -0

The tangential or shear stress on an oblique plane at an angle θ to the cross-section of a body which is subjected to a direct tensile stress (σ) is equal to

A
σ/2 sin 2θ

B
σ cos θ

C
σcos²θ

D
σsin²θ

Solution

We have \(\sigma=\frac{F}{A}\)
For an oblique plane we have \(\frac{F \cos \theta}{A / \cos \theta}=\frac{\sigma}{2} \sin 2 \theta\)
Hence, the correct option is (A)
Question 8
1 / -0

Two wires A and B are of the same materials. Their lengths are in the ratio 1: 2 and the
diameters are in the ratio \(2: 1,\) When stretched by force \(F_{A}\) and \(F_{B}\) respectively they get equal increase in their lengths. Then the ratio \(\frac{F_{A}}{F_{B}}\) should be:

A
1:2

B
1:1

C
2:1

D
8:1

Solution

Using Hooke's law \(Y=\frac{F / A}{\Delta L / L} \quad\) where \(\quad A=\frac{\pi D^{2}}{4}\)
\(\Longrightarrow F=Y \frac{\pi D^{2}}{4} \frac{\Delta L}{L}\)
As \(Y\) and \(\Delta L\) are constant, thus \(\quad F \propto \frac{D^{2}}{L}\) \(\Longrightarrow \frac{F_{A}}{F_{B}}=\frac{D_{A}^{2}}{D_{B}^{2}} \times \frac{L_{B}}{L_{A}}\)
Given: \(\quad D_{A}: D_{B}=2: 1\) and \(L_{A}: L_{B}=1: 2\)
\(\therefore \quad \frac{F_{A}}{F_{B}}=\frac{4}{1} \times \frac{2}{1}=\frac{8}{1}\)
Hence, the correct option is (D)
Question 9
1 / -0

The following substances which possess rigidity modulus

A
Only Solids

B
Only liquids

C
Liquids and Gases

D
Solids, Liquids and Gases

Solution

Only solids can sustain shear force.

Liquids and gases cannot sustain shear force.

Therefore, they don't possess rigidity modulus.
Hence, the correct option is (A)
Question 10
1 / -0

When water falls from a tap the velocity of falling water under the action of gravity will increase and the cross section of the water stream will decrease i.e., the falling stream of water becomes narrower.It is based on :

A
Principle of continuity

B
Bernoullis theorem

C
Law of conservation of energy

D
Law of conservation of linear momentum.

Solution

According to principle of continuity :

Av=constant

Thus decreasing the cross section area of the water stream will increase the velocity of water stream.
Hence, the correct option is (A)

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Physics Test - 27

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Physics Test - 27

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SHARING IS CARING

Question 1

No change in potential energy

Zero value of potential energy

Increase in potential energy

Decrease in potential energy

Solution

Question 2

0.5mm

1.0mm

0.05mm

0.06mm

Solution

Question 3

0.25

0.50

2.00

4.00

Solution

Question 4

L=200 cm,d=0.5 mm

L=300 cm,d=1.0 mm

L=50 cm,d=0.05 mm

L=100 cm,d=0.2 mm

Solution

Question 5

200 N

400 N

600 N

800 N

Solution

Question 6

Nearly 17 N

Nearly 34 N

Nearly 51 N

Nearly 68 N

Solution

Question 7

σ/2 sin 2θ

σ cos θ

σcos2θ

σsin2θ

Solution

Question 8

1:2

1:1

2:1

8:1

Solution

Question 9

Only Solids

Only liquids

Liquids and Gases

Solids, Liquids and Gases

Solution

Question 10

Principle of continuity

Bernoullis theorem

Law of conservation of energy

Law of conservation of linear momentum.

Solution

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σcos²θ

σsin²θ