Physics Test

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Physics Test - 18

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Question 1
4 / -1

Directions For Questions

Directions:The following question has four choices out of which ONLY ONE is correct.

Following question contains statements given in two columns, which have to be matched. The statements in Column I are labeled A, B, C and D, while the statements in Column II are labeled p, q, r, s and t. Any given statement in Column I can match with ONE OR MORE statements (s) in column II. The appropriate bubbles corresponding to the answers to these questions have to be darkened as illustrated in the following example:

...view full instructions

If the correct matches are A - p, s and t; B - q and r, C - p and q; and D - s and t: the correct darkening of bubbles will look the following.

A

B

C

D

Solution
Question 2
4 / -1

Starting with the same initial conditions, an ideal gas expands from volume V1 to V2 in three different ways. The work done by the gas is W1if the process is purely isothermal; W2, if purely isobaric, and W3, if purely adiabatic. Then:

A
W2> W1> W3

B
W2> W3> W1

C
W1> W2> W3

D
W1> W3> W2

Solution

The Work Done equals under the curve

W₂> W₁> W₃
Question 3
4 / -1

In the given figure, each plate of capacitance C has a partial value of charge equal to

A
CE

B
\(\frac{C E R_{1}}{R_{2}-r}\)

C
\(\frac{C E R_{2}}{R_{2}+r}\)

D
\(\frac{C E R_{1}}{R_{1}-r}\)

Solution

As capacitor behaves as an open circuit in \(\mathrm{dc}\), so the voltage across the capacitors = voltage across the resistance \(\mathrm{R}_{2}\)
Voltage across \(\mathrm{R}_{2}\) :
\(V=\left(\frac{R_{2}}{r+R_{2}}\right) E\)
Charge across the capacitor:
\(Q=C V\)
\(Q=\left(\frac{C R_{2}}{r+R_{2}}\right) E\)
Question 4
4 / -1

A stretched wire of same length under a tension vibrates with its fundamental frequency. Its length is decreased by 45% and tension is increased by 21%. Now, its fundamental frequency

A
increases by 50%

B
increases by 100%

C
decreases by 50%

D
decreases by 25%

Solution

The formula for fundamental frequency is \(\mathrm{n}=(1 / 2 \mathrm{l}) \mathrm{v}(\mathrm{T} / \mathrm{m}) .\) Where I is the
length, T is the Tension and \(\mathrm{m}\) is linear density. Here \(\mathrm{I}_{2}=\mathrm{I}_{1}(1+45 \%)\) and \(\mathrm{T}_{2}=\mathrm{T}_{1}(1+21 \%)\)
Now \(\mathrm{n}_{1} / \mathrm{n}_{2}=\left(\mathrm{l}_{2} / \mathrm{l}_{1}\right) \mathrm{v}\left(\mathrm{T}_{1} / \mathrm{T}_{2}\right) .\) Solving this we get 1: 2
Hence, the fundamental frequency increases by \(100 \%\)
Question 5
4 / -1

A man is standing on the platform. One train is approaching and another train is going away with speed of 4 m/s, frequency of sound produced by train is 240 Hz. What will be the number of beats heard by him per second?

A
12

B
Zero

C
6

D
3

Solution

For the approaching train, the frequency heard by the standing man is \(f_{1}\)
\(f_{1}=f\left(\frac{v}{v-v_{s}}\right)=240\left(\frac{300}{300-4}\right)\)
\(f_{1}=240\left(\frac{300}{296}\right)=243.24 \approx 243 H z\)
For the receding train, the frequency heard by the standing man is \(f_{2}\)
\(f_{2}=f\left(\frac{v}{v+v_{s}}\right)=240\left(\frac{300}{300+4}\right)\)
\(f_{2}=240\left(\frac{300}{304}\right)=236.68 \approx 237 \mathrm{Hz}\)
Hence, number of beats heard by him per second are
\(f_{1}-f_{2}=243-237=6\)
Question 6
4 / -1

A frog can be levitated in magnetic field produced by a current in a vertical solenoid placed below the frog. This is possible because the body of the frog behaves as a/an _______ substance.

A
paramagnetic

B
diamagnetic

C
ferromagnetic

D
anti-ferromagnetic

Solution

When diamagnetic substances are placed in a magnetic field, then they are feebly repelled in the field.

Frog is levitated in magnetic field produced by the current in vertical solenoid below the frog due to repulsion, so body of frog behaves as diamagnetic substance.
Question 7
4 / -1

A point mass 'm' is suspended at the end of a mass less wire of length 'L' and cross-section 'A'. If 'Y' is the Young's modulus for the wire, the frequency of oscillations for the SHM along the vertical line is

A
\(\frac{1}{2 \pi} \sqrt{\frac{\gamma A}{m L}}\)

B
\(2 \pi \sqrt{\frac{m L}{Y A}}\)

C
\(\frac{1}{\pi} \sqrt{\frac{Y A}{m L}}\)

D
\(\pi \sqrt{\frac{m L}{Y A}}\)

Solution

For stretched wire,
\(\frac{\text {Stress}}{\text {Strain}}=Y\)
\(\frac{\left(\frac{F}{A}\right)}{\left(\frac{l}{L}\right)}=Y\)
\(F=\frac{Y A}{L} l\)
Comparing this equation with
\(|F|=k y\)
Spring constant \(k=\frac{Y A}{L}\)
So, Frequency of oscillation will be
\(f=\frac{1}{2 \pi} \sqrt{\frac{k}{m}}\)
\(f=\frac{1}{2 \pi} \sqrt{\frac{Y A}{m L}}\)
Question 8
4 / -1

If a body is in equilibrium under the action of three co-planar forces, then;

A
they must act in a straight line

B
they must meet in a point or parallel

C
their horizontal and vertical components are equal

D
none of these

Solution

If a body is in equilibrium under the action of three co-planar forces, then they must meet in a point or parallel .
Question 9
4 / -1

If VB- VD= 0 in the given circuit, then what is the value of R?

A
3 ohms

B
4.5 ohms

C
18 ohms

D
28 ohms

Solution
Question 10
4 / -1

The spring constants of 3 springs connected to a mass are shown in the figure. When the mass oscillates, what is the time period of the oscillations?

A

B

C

D

Solution