Physics Test

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Physics Test - 19

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Question 1
4 / -1

A wave equation is given by $y=4 \sin \left[\pi\left(\frac{t}{5}-\frac{x}{9}+\frac{1}{6}\right)\right]$ where ${ }^{\prime} x^{\prime}$ is in $\mathrm{cm}$ and ${ }^{\prime} t^{\prime}$ in sec. Which of the following is true?

A
λ = 18 cm

B
'v' = 4 m/s

C
'A' = 0.4 m

D
'f' = 50 Hz

Solution

Comparing the given equation with standard equation
$y=A \sin \left[\frac{2 \pi}{T} t-\frac{2 \pi}{\lambda} t+\phi\right],$ we have
So amplitude of the wave is $A=4 \mathrm{cm}$
$\frac{2 \pi}{T}=\frac{\pi}{5} \Rightarrow \frac{1}{T}=f=0.1 \mathrm{Hz}$
$\frac{2 \pi}{\lambda}=\frac{\pi}{9} \Rightarrow \lambda=18 \mathrm{cm}$
Question 2
4 / -1

Two pendulums oscillate with the same amplitude but with a constant phase difference of 90×. If the maximum velocity of one is v, the maximum velocity of the other will be

A
V

B
2V

C
$\sqrt{2 v}$

D
$\frac{v}{\sqrt{2}}$

Solution

Because phase difference is constant, hence time periods must be equal i.e. velocities must be equal.
Question 3
4 / -1

If the resultant of two forces P and Q acting at an angle `α` is R, inclined at an angle `θ` with P, then cos θ =

A
(P + Q cos α) / P

B
(P + Q cos α) / Q

C
(P + Q cos α) / R

D
None of these

Solution

$\cos \theta=\left(\frac{P+Q \cos \alpha}{R}\right)$
Question 4
4 / -1

The graph below shows the variation of velocity (v) of a body with position (x) from the origin O.

Which of the following graphs correctly represents the variation of acceleration (a) with position (x)?

A

B

C

D

Solution
Question 5
4 / -1

If each of the two unlike parallel forces P and Q ( P > Q ) acting at a distance d apart be increased by S, then the point of application of the resultant is moved through a distance

A
$\frac{d}{P-Q}$

B
$\frac{S}{P-Q}$

C
$\frac{S d}{P-Q}$

D
$\frac{S}{d(P-Q)}$

Solution

Let $P$ and $Q$ be two unlike parallel forces acting at points $A$ and $B$ of a rigid body. Suppose the resultant
forces $\mathrm{P}-\mathrm{Q}$ acts at $\mathrm{C}$
Then,
\[
A C=\left(\frac{A B}{P-Q}\right) Q
\]

If each of the forces $P$ and $Q$ is increased by $s$, let the resultant act at $D$. Then, $A D=\left[\frac{A B}{(P+S)-(Q+S)}\right](Q+S)=\left[\frac{A B}{P-Q}\right](Q+S)$
Now
$C D=A D-A C$
$A D=\left(\frac{A B}{P-Q}\right)(Q+S)-\left[\frac{A B}{P-Q}\right] Q$
$=\left[\frac{A B}{P-Q}\right] S=\frac{d S}{P-Q}$
Question 6
4 / -1

An observer moves towards a stationary source of sound with a velocity one-fifth of the velocity of sound. What is the percentage increase in the apparent frequency?

A
Zero

B
0.5%

C
5%

D
None of these

Solution

$f^{\prime}=f\left(\frac{v-v_{0}}{v-v_{z}}\right)$
$f^{\prime}=f\left(\frac{v+v / 5}{v}\right)=\left(\frac{6 f}{5}\right)$
$\frac{f^{\prime}}{f}=\frac{6}{5} \Rightarrow \frac{\Delta f}{f}=\frac{1}{5}$
$\%$ change $=\frac{\Delta f}{f} \times 100=20 \%$
Question 7
4 / -1

A tube, closed at one end containing air, produces, when excited, the fundamental note of frequency 512 Hz. If the tube is open at both ends, then the fundamental frequency that can be excited will be (in Hz)

A
1024

B
512

C
256

D
128

Solution

For a closed tube $v=\frac{v}{4 L} .$ For an open tube $v^{\prime}=$ $\frac{v}{2 L} .$ Hence $v^{\prime}=2 v=2 \times 512=1024 \mathrm{Hz} .$ Thus the
correct choice is (1).
Question 8
4 / -1

The resultant of two forces P N and 3 N is a force of 7 N. If the direction of the 3 N force was reversed, the resultant would be $\sqrt{19}$ N. The value of P is

A
3 N

B
4 N

C
5 N

D
6 N

Solution

$7^{2}=p^{2}+3^{2}+2 P \times 3 \cos \alpha$
or $49=\mathrm{p}^{2}+9+6 \mathrm{P} \cos \alpha$
Again $19=\mathrm{P}^{2}+9-6 \mathrm{P} \cos \alpha$
(i) $+$ (ii) gives
$68=2 p^{2}+18 \quad p=5 N$
Question 9
4 / -1

An aircraft executes a horizontal loop at a speed of 720 km h-1with its wings banked at 150. What is radius of loop?

A
15.6 km

B
15.2 km

C
15.7 km

D
none of these

Solution

Given : Speed of air craft, $v=720 \mathrm{kmh}^{-1}$ $\begin{array}{ll}\Rightarrow & v=720 \times \frac{5}{18} \\ \Rightarrow & v=200 \mathrm{ms}^{-1}\end{array}$
Angle of banking, $\theta=15^{\circ}$ Radius of the loop $=r=?$ We know that for a banked curve,
$\begin{aligned} \tan \theta &=\frac{v^{2}}{r g} \\ \Rightarrow \quad r &=\frac{v^{2}}{g \tan \theta}=\frac{(200)^{2}}{9.8 \times \tan 15^{\circ}} \\ \Rightarrow \quad r &=\frac{40000}{9.8 \times 0.2679} \end{aligned}$
$=15.24 \times 10^{3} \mathrm{m}$
$\Rightarrow$
$r=15.24 \mathrm{km}$
Question 10
4 / -1

A circuit consists of five identical conductors as shown in figure. The two similar conductors are added as indicated by dotted lines.

The ratio of resistances before and after addition will be:

A
7/5

B
3/5

C
5/3

D
6/5

Solution

After adjoining two conductors, the circuit will acquire the form shown in the following figure. From symmetry considerations, we conclude that the central conductor will not be effective in electric charge transfer. Therefore, if the initial resistance $R_{1}$ of circuit was $5 \Omega,$ where resistance of each conductor is $1 \Omega,$ the new resistance of the circuit will become

$R_{2}=2+\frac{2}{2}=3 \Omega$
$\frac{R_{1}}{R_{2}}=\frac{5}{3}$