Physics Test

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Physics Test - 35

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Weekly Quiz Competition

Question 1
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A gaseous mixture consists of 16 of helium and 16 of oxygen. The ratio C_P / C_V of the mixture is

A
1.4

B
1.54

C
1.59

D
1.62

Solution

Mixture of non-reactive gases
(a) $n=n_{1}+n_{2}$
(b) $p=p_{1}+p_{2}$
(c) $\cup=U_{1}+U_{2}$
(d) $\Delta=\Delta U_{1}+\Delta U_{2}$
(e) $\mathrm{C}_{\mathrm{V}}=\frac{n_{1} \mathrm{C}_{\mathrm{v}_{1}}+n_{\mathrm{g}} \mathrm{C}_{\mathrm{v}_{2}}}{n_{1}+n_{2}}$
$(\mathrm{f}) \mathrm{C}_{\mathrm{p}}=\frac{n_{1} \mathrm{C}_{\mathrm{p}_{1}}+\mathrm{n}_{2} \mathrm{C}_{\mathrm{p}_{2}}}{\mathrm{n}_{1}+\mathrm{n}_{2}}=\mathrm{C}_{\mathrm{V}}+\mathrm{R}$
(g) $\gamma=\frac{c_{p}}{C_{v}}$ or $\frac{n}{\gamma-1}=\frac{n_{1}}{\gamma_{1}-1}+\frac{n_{2}}{\gamma_{2}-1}$
(h) $\mathrm{M}=\frac{n_{1} \mathrm{M}_{1}+m_{2} \mathrm{M}_{2}}{n_{1}+n_{2}}$
Question 2
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A solid sphere of radius R made of a material of bulk modulus B is surrounded by a liquid in a cylindrical container. A massless piston of area A floats on the surface of the liquid. Find the fractional change in the radius of the sphere (dR/R) when a mass M is placed on the piston to compress the liquid.

A
$\frac{\mathrm{Mg}}{2 \mathrm{AB}}$

B
$\frac{\mathrm{Mg}}{6 \mathrm{AB}}$

C
$\frac{\mathrm{Mg}}{3 \mathrm{AB}}$

D
$\frac{\mathrm{Mg}}{8 \mathrm{AB}}$

Solution

As for a spherical body, $V =\frac{4}{3} \pi R ^{3}, \quad \frac{\Delta R }{ R }=\frac{1}{3} \frac{\Delta V }{ V }$
Now by definition of bulk modulus, $B =- V \frac{\Delta p }{\Delta v }$ i.e., $\left|\frac{\Delta v }{ v }\right|=\frac{\Delta p}{ B }=\frac{ M g }{ A B } \quad\left[\right.$ as $\left.\Delta p =\frac{ M g }{ A }\right]$
$\frac{ d R }{ R }=\frac{ Mg }{3 AB }$

Hence, the correct option is (C)
Question 3
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Circuit for the measurement of resistance by potentiometer is shown. The galvanometer is first connected at point A and zero deflection is observed at length PJ = 10 cm. In second case it is connected at point C and zero deflection is observed at a length 30 cm from P. Then the unknown resistance X is

A
$\frac{R}{2}$

B
$\frac{R}{3}$

C
3 R

D
2 R

Solution

In potentiometer wire potential difference is directly proportional to length

Let potential drop unit length a potentiometer wire be $K$. For zero deflection the current will flow independently in two circuits
$IR = K \times 10$
In second case $IR + IX = K \times 30$
(2) $-(1)$ $\Rightarrow \quad X = k \times 20$
Divide (i) and (iii)
$\frac{R}{X}=\frac{1}{2}$
$\Rightarrow \quad x=2 R$

Hence, the correct option is (D)
Question 4
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One end of a light spring of natural length d and spring constant k is fixed on a rigid wall and the other is fixed to a smooth ring of mass m which can slide without friction in a vertical rod fixed at a distance d from the wall. Initially the spring makes an angle of 37^o with the horizontal as shown in diagram. When the system is released from rest, find the speed of the ring when the spring becomes horizontal. (sin 37^o = 3/5)

A
$\mathrm{d} \sqrt{\frac{3 \mathrm{g}}{2 \mathrm{d}}+\frac{\mathrm{k}}{16 \mathrm{m}}}$

B
$\mathrm{d} \sqrt{\frac{2 \mathrm{g}}{3 \mathrm{d}}-\frac{\mathrm{k}}{21 \mathrm{m}}}$

C
$\mathrm{d} \sqrt{-\frac{4 \mathrm{g}}{2 \mathrm{d}} \frac{\mathrm{k}}{12 \mathrm{m}}}$

D
$\mathrm{d} \sqrt{-\frac{8 \mathrm{g}}{4 \mathrm{d}}+\frac{\mathrm{k}}{28 \mathrm{m}}}$

Solution

If $l$ is the stretched length of the spring
$\frac{d}{l}=\cos 37^{\circ}=\frac{4}{5}, \quad$ i.e., $\quad l=\frac{5}{4} d$
So, the stretch $y=l-d=\frac{5}{4} d-d=\frac{d}{4}$
and
$A B = h =l \sin 37^{\circ}=\frac{5}{4} d \times \frac{3}{5}=\frac{3}{4} d$
Now taking point $B$ as reference level and applying law of conservation of Energy between $A$ and $B$
\begin{equation*}
E_{A}=E_{B}
\end{equation*}
$m g h+\frac{1}{2} k y^{2}+0=0+0+\frac{1}{2} m v^{2}$
$[$ as for $B , h =0$ and $y =0]$
or $\frac{3}{4} m g d+\frac{1}{2} k\left(\frac{d}{4}\right)^{2}=\frac{1}{2} m v^{2}$
$\left[\right.$ as for $A, h=\frac{3}{4} d$ and $\left.y=\frac{1}{4} d\right]$ or $v = d \sqrt{\frac{3 g }{2 d }+\frac{ k }{16 m }}$

Hence, the correct option is (A)
Question 5
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A small cannon 'A' is mounted on a platform ( which can be rotated so that the cannon can aim at any point) and is adjusted for maximum range R of shells. It can throw shells on any point on the shown circle (dotted) on ground. Suddenly a windstorm starts blowing in horizontal direction normal to AB with a speed $\sqrt{2}$ times the velocity of shell. At what least distance can the shell land from point B.(Assume that the velocity of the windstorm is imparted to the shell in addition to its velocity of projection. Also assume that the platform is kept stationary while projecting the shell.)

A
$\mathrm{R}(\sqrt{5}-1)$

B
$\mathrm{R}(\sqrt{3}-1)$

C
$R \sqrt{3}$

D
$R \sqrt{5}$

Solution

Let the speed of shell be u and the speed of wind be $v$ The time of flight T remains unchanged due to windstorm, $\mathbf{T}=\frac{\sqrt{2} \mathrm{u}}{\mathrm{g}}$
(1)
Horizontal component of velocity of shell in absence of air, $\mathrm{u}_{\mathrm{H}}=\frac{\mathrm{u}}{\sqrt{2}}$
(2)
Hence the net $x$ and $y$ component of velocity of shell (see figure) are $\mathrm{u}_{x}=\sqrt{2} \mathrm{u}+\frac{\mathrm{u}}{\sqrt{2}} \cos \theta$
(3a) $u_{y}=\frac{n}{\sqrt{2}} \sin \theta$
(3b)
$\therefore$ The $\mathrm{x}$ and $\mathrm{y}$ coordinate of point $\mathrm{P}$ where shell lands is
$x=u_{x} T=\left(\sqrt{2} u+\frac{u}{\sqrt{2}} \cos \theta\right) \frac{\sqrt{2} u}{g}=2 R+R \cos \theta$
$y=u_{y} T=\left(\frac{u}{\sqrt{2}} \sin \theta\right) \frac{\sqrt{2} u}{8}=R \sin \theta$
$\therefore$ The distance S between $B$ and $P$ is given by.
$\mathrm{S}^{2}=(x-0)^{2}+(y-\mathrm{R})^{2}=(2 \mathrm{R}+\mathrm{R} \cos \theta)^{2}+(\mathrm{R} \sin \theta-\mathrm{R})^{2}$
$=\mathrm{R}^{2}[6+4 \cos \theta-2 \sin \theta]$
$=\mathrm{R}^{2}\left[6+\sqrt{20}\left(\frac{4 \cos \theta}{\sqrt{20}}-\frac{2 \sin \theta}{\sqrt{20}}\right]\right.$
$\therefore \mathrm{S}_{\text {minimum }}=\mathrm{R} \sqrt{6-\sqrt{20}}=\mathrm{R} \sqrt{6-2 \sqrt{5}}$ or $\mathrm{R}(\sqrt{5}-1)$
Hence, the correct option is (A)
Question 6
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A thread is wound around two discs on either sides. The pulley and the two discs have the same mass and radius. There is no slipping at the pulley and no friction at the hinge. Find out the accelerations of the two discs and the angular acceleration of the pulley

A
$\alpha=0, a_{1}=a_{2}=\frac{2 \mathrm{g}}{3}$

B
$\alpha=1, a_{1}=a_{2}=\frac{2 g}{5}$

C
$\alpha=0, \alpha_{1}=\alpha_{=} \frac{2 \mathrm{g}}{1}$

D
$\alpha=0, \alpha_{1}=\alpha_{2}=\mathrm{g}$

Solution

Let $R$ be the radius of the discs and $T_{1}$ and $T_{2}$ be the tensions in the left and right segments of the rope.

Acceleration of disc 1
$a _{1}=\frac{ m g- T _{1}}{m}$
Acceleration of disc 2 ,
$a _{2}=\frac{ mg -T_{2}}{m}$
Angular acceleration of disc 1 ,
$\alpha_{1}=\frac{\tau}{I}=\frac{T_{1} R}{\frac{1}{2} m R^{2}}=\frac{2 T_{1}}{m R}$
Similarly, angular acceleration of disc
2. $\quad \alpha_{2}=\frac{2 T_{2}}{ mR }$
Both $a_{1}$ and $a_{2}$ are clockwise. Angular acceleration of pulley, $\alpha =\frac{\left(T_{2}-T_{1}\right) R}{\frac{1}{2} m R^{2}}=\frac{2\left(T_{2}-T_{1}\right)}{m R} \quad \ldots(v)$
For no slipping. $Ra _{1}- a _{1}= a _{2}- Ra _{2}= Ra$

Solving these equations, we get $\alpha = 0$
and $a_{1}=a_{2}=\frac{2 \pi}{3}$
Alternate solution As both the discs are in identical situation, $T_{1}=T_{2}$ and $a=0$.
i.e., each of the discs fails independently and identically. Therefore, this is exactly similar to the problem shown in figure.

Hence, the correct option is (A)
Question 7
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In the shown circuit the resistance R can be varied:

The variation of current through R against R is correctly plotted as :

A

B

C

D

Solution

Redrawing the given circuit diagram as shown below:
Let $\vee$ be the potential difference between two points as shown,
i.e. $\mathrm{I}_{1}+\mathrm{I}_{2}+\mathrm{I}_{3}=0$
Using point potential theory,

$\begin{array}{ll} & \frac{\mathrm{V}-\mathrm{E}}{\mathrm{r}}+\frac{\mathrm{V}-\mathrm{E}}{\mathrm{r}}+\frac{\mathrm{V}-\mathrm{E}}{\mathrm{R}}=0 \\ \Rightarrow & (\mathrm{V}-\mathrm{E})\left(\frac{2}{\mathrm{r}}+\frac{1}{\mathrm{R}}\right)=0 \\ \text { As } & \frac{2}{\mathrm{r}}+\frac{1}{\mathrm{R}} \neq 0 \\ \text { So } & \mathrm{V}-\mathrm{E}=0\end{array}$
So, current through $\mathrm{R}$,
$\mathrm{i}=\frac{\mathrm{V}-\mathrm{E}}{\mathrm{R}}=0$ whatever be the value of $\mathrm{R}$
Hence, the correct option is (D)
Question 8
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A projectile is fired into the air from the edge of a 100 m high cliff at an angle of 37⁰ above the horizontal. The projectile hits a target 400 m away from the base of the cliff. What is the initial velocity of the projectile, v_o ? ( Neglect air friction and assume x- axis to be horizontal and y - axis to be vertical).

A
$\mathrm{v}_{0}=25 \sqrt{5} \mathrm{m} / \mathrm{s}$.

B
$\mathrm{v}_{0}=23 \sqrt{3} \mathrm{m} / \mathrm{s}$.

C
$\mathrm{v}_{0}=22 \sqrt{2} \mathrm{m} / \mathrm{s}$.

D
$\mathrm{v}_{0}=27 \sqrt{7} \mathrm{m} / \mathrm{s}$.

Solution

Let t be the time after which projectile reaches the ground, motion in horizontal directic
$400=\left( v _{0} \cos 37^{0}\right) t$
$400= v _{0}(4 / 5) t$
$\Rightarrow \quad v_{0} t=500 \ldots \ldots \ldots . \ldots \ldots .(1)$
Taking motion in vertical direction,
$h = u t + 1 / 2 g t ^{2}$
$\Rightarrow \quad 100=\left(-v_{0} \sin 37^{0}\right) t+1 / 2(10) t^{2}$
$\Rightarrow 100=-\frac{3}{5}\left(v_{0} t\right)+5 t^{2}$
Putting $v_{0} t=500$ From equation (1) $\Rightarrow 100=-\frac{3}{5}(500)+5 t^{2}$
$\Rightarrow 5 t^{2}=400 \Rightarrow t=\frac{20}{\sqrt{5}} \sec$
From eqn (1)$; v_{0}=500 \times \frac{\sqrt{5}}{20} \sec$
$v _{0}=25 \sqrt{5} m / s$

Hence, the correct option is (A)
Question 9
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A piece of copper and another of germanium are cooled from room temperature to 80 K. The resistance of

A
Each of them increases

B
Each of them decreases

C
Copper increases and germanium decreases

D
Copper decreases and germanium increases

Solution

Copper is metal and germanium is semiconductor. Resistance of a metal decreases and that of a semiconductor increases with decrease in temperature.

KEY CONCEPTS
Temperature Dependence of Resistivity and Resistance $\rho( T )=\rho_{0}\left[1+\alpha\left( T - T _{0}\right)\right]$
where, $\rho_{0}$ is the resistivity at a reference temperature $T_{0}$ (often taken as $0^{\circ} C$ or $20^{\circ} C$ ) and $\rho( T )$ is the resistivity at temperature T, which may be higher or lower than To. The factor $\alpha$ is called the temperature coefficient of resistivity.
$R ( T )= R _{0}\left[ 1 + \alpha \left( T - T _{0}\right)\right]$

Hence, the correct option is (D)
Question 10
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A ball of mass m moving with velocity v strikes the bob of a pendulum at rest. The mass of the bob is also m. If the collision is perfectly inelastic, the height to which the bob will rise is given by

A
$\frac{v^{2}}{8 g}$

B
$\frac{v^{2}}{2 g}$

C
$\frac{v^{2}}{6 g}$

D
$\frac{v^{2}}{4 g}$

Solution

Because the collision is perfectly inelastic, hence the two blocks stick together. By conservation of linear momentum,

2Mv=MV

By conservation of energy,

$2 \mathrm{mgh}=\frac{1}{2} 2 \mathrm{mV}^{2}$ or $\mathrm{h}=\mathrm{v}^{2} / 8 \mathrm{g}$

Concepts :
Main Concept :
Conservation of Linear MomentumWhen the sum of the forces on an object is zero, the equation
$\overrightarrow{\mathrm{F}}_{\text {net }}=\frac{\mathrm{d} \overrightarrow{\mathrm{p}}}{\mathrm{dt}}$
tell us that the time derivative of momentum is zero.

that is, $\frac{\mathrm{d} \overrightarrow{\mathrm{p}}}{\mathrm{dt}}=0$

this implies $\overrightarrow{\mathrm{p}}=$ constantt
Consequently, one can state a conservation law for momentum:
When the net impulse acting on a particle is zero, its momentum is constant.
The real utility of the momentum conservation concept comes about when it is applied to a collection of particles. For a system of particles, the total momentum is simply the vector sum of the momentum of each of the particles in the system. That is
$\overrightarrow{\mathrm{P}}=\sum_{i=1}^{N} \overrightarrow{\mathrm{p}}_{\mathrm{i}}$
Now consider the net force on a system of particles.
There are two kinds of forces:
(i) Internal forces, resulting from the interaction between the particles within the system, and
(ii) External forces, arising from the interaction between the particles in the system and objects outside the system.
For example, consider a system of two blocks joined together with a spring. If the system is allowed to fall freely under gravity, then the gravitational force acting on each block is the external force and the spring force acting on each block is internal.
When we calculate the net force on a system of particles by performing the vector sum, then the summation of all the internal forces is always zero. If the summation of all the external forces is also zero, then the net impulse acting on the body is also zero. Hence, momentum of the system is conserved.
i.e. P→sys= constant
Application of Conservation of Momentum
The conservation of momentum law can be used to relate the initial motion of particles within a system to the motion of those same particles sometime later. The law emphasizes the equality of momentum before and after something happens within the system.
Thus, the conservation law is usually written as
${P →}_{initial} = {P →}_{final}$
$\sum_{i = 1}^{N} {p →}_{i} = \sum_{f = 1}^{N} {p →}_{f}$
Hence, the correct option is (A)