Physics Test

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Physics Test - 45

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Question 1
1 / -0

An 80 kg man standing on ice throws a 4 kg body horizontally at 6 ms⁻¹. The frictional coefficient between the ice and his feet is 0.02. The distance through which the man slips is (g = 10 ms⁻²)

A
0.225 m

B
2.25 m

C
0.115 m

D
0.20 m

Solution

While throwing the body, velocity gained by the man is $V_{x}=\frac{4 \times 6}{80}=0.3 \mathrm{ms}^{-1}$
Stopping distance $S=\frac{V^{2}}{2 \mu g}$
$s=\frac{0.3 \times 0.3}{2 \times 0.02 \times 10}$
$s=\frac{9}{40}=0.225 m$
Hence, the correct option is (A)
Question 2
1 / -0

A spring mass system with unequal masses is placed at rest on a smooth horizontal surface. The spring is initially kept compressed with a thread. When the spring is cut, the mass m moves so as to enter in a vertical circular loop of radius r. The minimum compression x in the spring so that the mass m may negotiate the vertical loop is

A
$\frac{5 m g r}{k} \cdot \frac{m}{M}$

B
$\frac{3 m g r}{k} \cdot \frac{M-m}{m}$

C
$\sqrt{\frac{5 m g r}{k}\left(\frac{M+m}{M}\right)}$

D
$\sqrt{\frac{3 m g r}{k} \cdot \frac{M}{m}}$

Solution

Let $x$ be the compression in the spring, $v$ and $V$ be the velocities of $m$ and $M$ respectively, when the spring released.
According to the law of conservation of momentum and energy $M V=m v \ldots(1)$
$a n d \frac{1}{2} k x^{2}=\frac{1}{2} M V^{2}+\frac{1}{2} m v^{2} \ldots .(2)$
From (2) and (1) $k x^{2}=\frac{m^{2} v^{2}}{M}+m v^{2}$
$\Rightarrow k x^{2}=\left(\frac{m+M}{M}\right) m v^{2}$
$\therefore x=v \sqrt{\frac{(M+m)}{M} \frac{m}{k}}$
To negotiate a loop, minimum velocity required $v=\sqrt{5 g R}$
$\therefore x=\sqrt{\frac{5 m g r}{k}\left(\frac{M+m}{M}\right)}$
Hence, the correct option is (C)
Question 3
1 / -0

A satellite orbiting around the earth of radius R is shifted to an orbit of radius 2R. The time taken for one revolution will increase nearly by

A
2 times

B
2.5 times

C
2.8 times

D
3.2 times

Solution

The time period of a satellite orbiting around the earth is $T=2 \pi \sqrt{\frac{R^{3}}{G M}}$
When the radius of the orbit is increased to $2 \mathrm{R}$ $T^{\prime}=2 \pi \sqrt{\frac{(2 R)^{3}}{G M}}$
$\therefore \frac{T^{\prime}}{T}=\sqrt{\frac{8 R^{3}}{R^{3}}}=\sqrt{8}=2 \sqrt{2}$
$\therefore T^{\prime}=2 \sqrt{2} T$
$T=2 \times 1.414 T$
$T^{\prime}=2.828 T \approx 2.8 T$
Hence, the correct option is (C)
Question 4
1 / -0

Four metallic plates each with a surface area A of each side are placed at a distance d from each other. When the plates are connected as shown in Figure, the capacitance of the whole system between a and b is

A
$\frac{3 \varepsilon_{0} A}{2 d}$

B
$\frac{2 \varepsilon_{0} A}{3 d}$

C
$\frac{2 \varepsilon_{0} A}{d}$

D
$\frac{3 \varepsilon_{0} A}{d}$

Solution
Question 5
1 / -0

A battery of internal resistance 4 Ω is connected to the network of resistance as shown in the Figure. In order to get maximum power to the network, the value of R in ohm should be

A
1 Ω

B
1.5 Ω

C
2 Ω

D
2.5 Ω

Solution
Question 6
1 / -0

Steam at 100 °C is passed into 1.4 kg of water kept in a calorimeter of water equivalent 0.03 kg at 20 °C till the temperature of the calorimeter and contents reach 80 °C. Latent heat of steam is 2.26 × 10⁶ J kg ^{− 1} and sp.heat capacity of water is 4200 J /kg/ °C. The mass of steam condensed in kilogram is nearly equal to

A
0.15 kg

B
0.16 kg

C
0.18 kg

D
0.20 kg

Solution

(M + Mw) C(80-20) = mL + mC(100 − 80)

(1.4 + 0.03) 4200 (80 − 20)= m × 2.26 × 106 + m × 4200 × 20

1.43 × 4200 × 60= m (2.26 × 106 + 4200 × 20)

$m=\frac{1.43 \times 4200 \times 60}{2.26 \times 10^{6}+84000}$

$=\frac{360360}{2344 \times 10^{3}}$

= 0.1537 kg ≈ 0.15 kg
Hence, the correct option is (A)
Question 7
1 / -0

An object is projected up the incline with a velocity of 24 m/s at angle of 30 ° as shown in the Figure. The distance up the inclined plane at which the object strikes is (g = 10 ms ^{− 2} )

A
32.6 m

B
34.2 m

C
36.4 m

D
38.4 m

Solution
Question 8
1 / -0

The given graph shows the extension ( Δl ) of a wire of length 1.0 m suspended from the top of a roof at one end and loaded at the other end. If the cross sectional area of the wire is 10 ^{− 6} m², the Young’s modulus Y of the material of the wire is

A
2 × 10 ^{− 7} Nm ^{− 2}

B
3 × 10⁷ Nm ^{− 2}

C
2 × 10¹¹ Nm ^{− 2}

D
2 × 10⁹ Nm ^{− 2}

Solution
Question 9
1 / -0

A point mass 0.1 kg executes S.H.M of amplitude 0.2 m. When the particle passes through the mean position, its kinetic energy is 9 × 10 ^{− 3} J. The equation of motion of their particle is S.H.M when the initial phase 60 ° is

A
$0.2 \sin \left(\pm 4 t+\frac{\pi}{2}\right)$

B
$0.1 \sin \left(\pm 6 t+\frac{\pi}{3}\right)$

C
$0.1 \sin \left(\pm 2 t+\frac{2 \pi}{3}\right)$

D
$y=0.2 \sin \left(\pm \frac{3}{\sqrt{2}} t+\frac{\pi}{3}\right)$

Solution

The displacement of a particle in S.H.M is $y=a \sin (\omega t+\varphi)$
Velocity $\frac{d y}{d t}=\omega \cos (\omega t+\varphi)$
The velocity is maximum when the particle passes through the mean position is $\left(\frac{d y}{d t}\right)_{\max }=\omega a$
The kinetic energy at this instant is $1 / 2(m)(\omega a)^{2}=9 \times 10^{-3} J$
$\frac{1}{2} \times 0.1 \times \omega^{2} \times(0.2)^{2}=9 \times 10^{-3}$
$\omega^{2}=\frac{9}{2}$
$\omega=\pm \frac{3}{\sqrt{2}}$
Putting the values of $a, \omega$ and $\varphi$ in the S.H.M. equation $\mathrm{y}=0.2 \sin \left(\pm \frac{3}{\sqrt{2}} \mathrm{t}+\frac{\pi}{3}\right)$
Hence, the correct option is (D)
Question 10
1 / -0

A cylinder of radius r = 1 m and height 3 m filled with a liquid upto 2 m high and is rotated about its vertical axis as shown in the Figure.

The speed of rotation of the cylinder when the point at the centre of the base is just exposed is

A
$\sqrt{48} r a d s^{-1}$

B
$\sqrt{54} r a d s^{-1}$

C
$\sqrt{60} r a d s^{-1}$

D
$$\sqrt{66} \mathrm{rads}^{-1}$$

Solution