Physics Test

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Physics Test - 18

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Question 1
1 / -0

A uniform solid sphere of radius R and mass M purely rolls down an inclined plane. The coefficient of friction between the sphere and the inclined plane is μ. The maximum value of θ for this to be possible is : (Take g = 10 m/s2)

A
$\tan ^{-1} \mu$

B
$\tan ^{-1}\left(\frac{7 \mu}{2}\right)$

C
$\tan ^{-1}\left(\frac{5 \mu}{3}\right)$

D
$\tan ^{-1}\left(\frac{3 \mu}{2}\right)$

Solution

Let friction 'f' act on solid sphere in upward direction-
then, $\mathrm{mg} \sin \theta-\mathrm{f}=\mathrm{ma}-$. (i)
and $f . R=l \alpha \ldots$ (ii)
As sphere is in pure rolling motion,
$\ldots(i i)$
Putting value of ' $\alpha^{\prime}$ in equation (ii)
f. $R=\left(\frac{2}{5} m R^{2}\right)\left(\frac{a}{R}\right)$
$\Rightarrow \mathrm{f}=\frac{2}{5} \mathrm{ma}$
Form equation (i) $\mathrm{mg} \sin \theta-2 / 5 \mathrm{ma}=\mathrm{ma}$
$\Rightarrow \quad a=\frac{5}{7} g \sin \theta$
and $\quad \alpha=\frac{5}{7 \mathrm{R}} \mathrm{g} \sin \theta$
Now, maximum friction that can be act on sphere $\mathrm{f}_{\max }=\mu \mathrm{N} \mathrm{f}_{\max }=\mu \mathrm{mg} \cos \theta$
So, $\mathrm{f}$ should be less than $\mathrm{f}_{\max } \Rightarrow \mathrm{f}<\mathrm{f}_{\max }$
$\Rightarrow \frac{2}{5} \mathrm{ma}<\mu \mathrm{mg} \cos \theta$
$\Rightarrow \frac{2}{5} \mathrm{m}\left(\frac{5}{7} \mathrm{g} \sin \theta\right)<\mu \mathrm{mg} \cos \theta$
$\Rightarrow \theta \leq \tan ^{-1}\left(\frac{7 \mu}{2}\right)$
Hence, the correct option is B.
Question 2
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A spring mass system is placed on a frictionless horizontal surface as shown in the figure. The spring is expanded by 1/10m and the blocks are given velocities as shown, then maximum extension of spring is :

A
$\frac{1}{10 \sqrt{2} \mathrm{m}}$

B
$\frac{1}{5 \sqrt{2}} \mathrm{m}$

C
$\frac{1}{20} \mathrm{m}$

D
$\frac{1}{10} m$

Solution

As there is no loss of energy Initial mechanical energy = Final mechanical energy For initial Mechanical energy
$M . E_{\text {inital }}=\frac{1}{2} k x^{2}+\frac{1}{2} m\left(v_{\text {relative }}\right)^{2}$
Where $\mu=\frac{m_{1} m_{2}}{m_{1}+m_{2}}=2$ and $v_{\text {relative }}=5 \mathrm{m} / \mathrm{s}$
$\mathrm{M} \cdot \mathrm{E}_{\text {inital }}=\frac{1}{2}(5000)\left(\frac{1}{10}\right)^{2}+\frac{1}{2}(2)(5)^{2}$
$=25+25=50$
For final Mechanical energy When there is maximum extension relative velocity between block $=0$ $M \cdot E_{\text {final }}=\frac{1}{2} k\left(x_{\max }\right)^{2}$
So, $\frac{1}{2} \mathbf{k}\left(\mathbf{x}_{\max }\right)^{2}=50$
$\Rightarrow 5000\left(x_{\max }\right)^{2}=100$
$\Rightarrow x_{\max }=\frac{1}{\sqrt{50}}=\frac{1}{5 \sqrt{2}} \mathrm{m}$
Hence, the correct option is B.
Question 3
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An open water tanker moving on a horizontal straight road has a cubical block of cork floating over its surface. If the tanker has an acceleration of ‘a’ as shown, the acceleration of the cork w.r.t container is (ignore viscosity) –

A
Zero

B

$\frac{a^{2}}{g}$

C

$\frac{a}{g} \sqrt{g^{2}-a^{2}} $

D
a

Solution
Question 4
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A man of mass 60 kg standing on a platform executing S.H.M. in the vertical plane. The displacement from the mean position varies as y = 0.5 sin (2π ft). The value of f, for which the man will feel weightlessness at the highest point is : (y is in metres)

A
$\frac{g}{4 \pi}$

B
$4 \pi \mathrm{g}$

C
$\frac{\sqrt{2 \mathrm{g}}}{2 \pi}$

D
$2 \pi \sqrt{2 g}$

Solution

weightlessness will be felt when acceleration of object towards earth = g
So, at highest point acceleration of plate = g
In S.H.M., y = 0.5 sin (2π ft)
acceleration at extreme point a = w2A
g = 0.5 (4π2f2)
$f=\frac{\sqrt{2 g}}{2 \pi}$
Hence, the correct option is C.
Question 5
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A silicon diode is connected in series with resistance and battery. What should be the value of battery if the reading of ammeter 3.62 A.

A
25.6

B
40.0

C
44.1

D
51.5

Solution

Rearranging the circuit,

As we know the potential drop across silicon diode in forward biased is through the circuit is $I=\frac{\Delta V}{R}$ $3.62=\frac{V-0.7}{\frac{640}{59}}$

$\frac{3.62 \times 640}{59}=V-0.7$

$V-0.7=39.2$

$V=40 V$
Hence, the correct option is B.
Question 6
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A carrier wave has power of 1675 kW. If the side band power of a modulated wave subjected to 60%. Then find the amplitude modulation level

A
301.5KW

B
942KW

C
1392KW

D
1884KW

Solution

As you know that, sideband power is, $P_{S}=\frac{1}{2} m^{2} P_{C}$, Where $P_{C}$ is power of carrier wave, $P_{S}$ is sideband power and 'm' is modulation rate, $P_{C}=\frac{2 P_{S}}{m^{2}}$
$P_{S}=\frac{m^{2} P_{C}}{2}$
$P_{S}=\frac{0.6 \times 0.6 \times 1675}{2}$
$P_{S}=301.5 \mathrm{kW}$
Hence, the correct option is A.
Question 7
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A particle is moving eastwards with a velocity of 5 ms−1. In 10 seconds the velocity changes to 5 ms−1 northwards. The average acceleration in this time is

A
1/√ 2ms⁻²N-W

B
1/2ms⁻² N

C
Zero

D
1/2ms⁻² N-E

Solution

$v_{A}=5 \hat{i}$
$v_{B}=5 \hat{j}$
Change in velocity $=5 \hat{j}-5 \hat{i}$
$A v g-a c c^{n}=\frac{\text { change in velocity }}{\text { time }}=\frac{5 \hat{j}-5 \hat{i}}{10}=\frac{\hat{j}-\bar{i}}{2}$
Magnitude of this vector $=\sqrt{\frac{1^{2}+1^{2}}{4}}=\frac{1}{\sqrt{2}}$
Direction of this vector is $\tan \theta=\frac{-1}{1} \Rightarrow \theta=135^{\circ},$ i.e, N-W direction
Hence, the correct option is A.
Question 8
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A particle move along the parabolic path x=y2+2y+2 in such a way that the y-component of velocity vector remains 5m/s during the motion. The magnitude of the acceleration of the particle is:

A
50m/s²

B
100m/s²

C
102√m/s²

D
0.1m/s²

Solution

$x=y^{2}+2 y+2$
$v_{x}=\frac{d x}{d t}=2 y \frac{d y}{d t}+2 \frac{d y}{d t}=2 v_{y}(y+1)$
since the $y$ component of velocity remain the same, there is no acceleration along the y component, $a_{y}=0$ $a_{x}=\frac{d v_{x}}{d t}=2 x_{y}\left(\frac{d y}{d t}\right)=2 v_{y} v_{y}=2 \times 5 \times 5=50 \mathrm{ms}^{-2}$
Hence, the correct option is A.
Question 9
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Mass A is released from rest at the top of a frictionless inclined plane 18m long and reaches the bottom 3s later. At the instant when A is released, a second mass B is projected upwards along the plate from the bottom with a certain initial velocity. Mass B travels a distance up the plane, stops and returns to the bottom so that it arrives simultaneously with A. The two masses do not collide. Initial velocity of B is

A
4ms−1

B
5ms−1

C
6ms−1

D
7ms−1

Solution

Here, for $A, 18=0 \times 3+\frac{1}{2} a \times 3^{2}$ or $a=4 m s^{-2}$
For $B$, time taken to move up is given by, $t_{1}=u / a(\therefore$, the relation $v=u+a t$, here becomes $0=u-a t_{1}$ ). Distance moved up is given by the relation $0=u^{2}-2 a S$ i.e $S=u^{2} / 2 a$
For coming down the inclined plane for $S=\frac{1}{2} a t_{2}^{2}$
and $t_{2}=\sqrt{\frac{2 S}{a}}$
or $t_{2}=\sqrt{\frac{2 u^{2}}{a .2 a}}=\frac{u}{a}$
But $\frac{u}{a}=t_{1},$ then
$t_{1}+t_{2}=3$ or $\frac{2 u}{a}=3$ ot
$u=\frac{3 a}{2}$ or
$u=\frac{3}{2} \times 4=6 m s^{-1}$
Hence, the correct option is C.
Question 10
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A particle moving with velocity equal to 0.4 m/s is subjected to an acceleration of 0.15m/s² for 2 s in a direction at right angles to its direction of motion. The magnitude of resultant velocity is

A
0.3 m/s

B
0.5 m/s

C
0.27 m/s

D
0.35 m/s

Solution

$v_{x}=0.4 \mathrm{m} / \mathrm{s}$
$v_{y}=0 \mathrm{m} / \mathrm{s}$
$a_{y}=0.15 \mathrm{m} / \mathrm{s}^{2}, t=2 \mathrm{sec}$
$\left(v_{y}\right)_{\text {final }}=v_{y}+a_{y} t=0+(0.15) \times 2=0.3 \mathrm{m} / \mathrm{s}$
So, $v=v_{x} \hat{i}+v_{y} \hat{j}=0.4 \hat{i}+0.3 \hat{j}$
$v_{\text {nre }}=|v|=\sqrt{(0.4)^{2}+(0.3)^{2}}=0.5 \mathrm{m} / \mathrm{s}$
Hence, the correct option is B.