Physics Test

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Physics Test - 18

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Weekly Quiz Competition

Question 1
1 / -0

A particle of mass m is moving in a circular path of constant radius r such that its centripetal acceleration a_c is varying with time t as a_c=k² rt², where k is a constant. The power delivered to the particle by the force acting on it is

A
2π mk² r²

B
mk² r²t

C
$\frac{(m K^{4} r^{2} t^{5})}{5}$

D
Zero

Solution

ac=k² rt²

or $\frac{v^{2}}{e} = k^{2} r t^{2}$

or v = krt

Therefore, tangential acceleration, a_t = $\frac{d v}{d t}$ = kr

or Tangential force,

F_t= ma_t= mkr

Only tangential force does work.

Power = F_t v = (mkr)(krt)

or Power=mk² r² t
Question 2
1 / -0

A conducting wire frame is placed in a magnetic field, which is directed into the paper, figure. The magnetic field is increasing at a constant rate. The directions of induced current in wires AB and CD are

A
A to B and C to D

B
B to A and C to D

C
A to B and D to C

D
B to A and D to C

Solution

As the magnetic field directed into the paper is increasing at a constant rate, therefore, induced current should produce a magnetic field directed out of the paper. Thus current in both the loops must be anti-clock-wise.

As area of loop on right side is more, therefore, induced emf on right side of loop will be more compared to the emf induced on the left-side of the loop
$[∴ e = \frac{d ϕ}{d t} = - A \frac{d B}{d t}]$

Hence net current will flow from D to C and B to A
Question 3
1 / -0

An artificial satellite is moving in a circular orbit around the earth with a speed equal to half the magnitude of escape velocity from the earth. The height of satellite above the surface of earth is -(where R=radius of earth)

A
R

B
2 R

C
3 R

D
4 R

Solution

escape velocity $V_{s} = \sqrt{\frac{2 G m}{R}}$

orbital velocity $V_{0} = \sqrt{\frac{G m}{r}}$

given $V_{0} = \frac{1}{2} V_{e}$

⇒ $\sqrt{\frac{G m}{R + h}} = \frac{1}{2} \sqrt{\frac{2 G m}{R}}$

⇒ R+h=2R or h=R
Question 4
1 / -0

A ball of mass m₁ is moving with velocity 3v. It collides head on elastically with a stationary ball of mass m₂. The velocity of both the balls become v after collision. Then the value of the ratio $\frac{m_{2}}{m_{1}}$ is

A
1

B
2

C
3

D
4

Solution

3m₁v = m₁v + m₂v

3m₁v − m₁v = m₂v

2m₁v = m₂v

∴ $\frac{m_{2}}{m_{1}} = 2$
Question 5
1 / -0

The ratio of work done by an ideal monoatomic gas to the heat supplied to it in an isobaric process is

A
$\frac{2}{5}$

B
$\frac{3}{2}$

C
$\frac{3}{5}$

D
$\frac{2}{3}$

Solution

$\eta=\frac{\mathrm{w}}{\mathrm{Q}}=\frac{\mathrm{R}}{\mathrm{Cp}}=\frac{\mathrm{R}}{\frac{5 \mathrm{R}}{2}}=\frac{2}{5}$
Question 6
1 / -0

Figure shows a smooth inclined plane of inclination θ fixed in a car. A sphere is set in pure rolling on the incline. For what value of 'a' (the acceleration of car in horizontal direction) the sphere will continue pure rolling?

A
g cos θ

B
g sin θ

C
g cot θ

D
g tan θ

Solution

The sphere will continue pure rolling if

ma cos θ = mg sin θ

or a =g tan θ
Question 7
1 / -0

While using travelling microscope in an experiment one find that one main scale division length is 1 mm and 10 VSD = 9 MSD. When this microscope is focused over a glass slab which is placed over a spot. Then which of the reading is correctly written -

A
1.6 cm

B
1.62 cm

C
1.622 cm

D
2 cm

Solution

LC= $\frac{1 m m}{10}$ =0.1 mm=0.01 cm

∴ Reading should be written upto 2 places after decimal (in cm)
Question 8
1 / -0

Two bodies of masses 10 kg and 100 kg are separated by a distance of 2 m. The gravitational potential at the mid-point of the line joining the two bodies is:

A
−7.3×10⁻⁷ J/k

B
−7.3×10⁻⁸ J/kg-7.3×10-8 J/kg

C
−7.3×10⁻⁹ J/k

D
−7.3×10⁻⁶ J/kg

Solution

As we know, gravitational potential
$V = \frac{- G M}{R}$
Gravitational potential at mid point,

$=-\frac{G \times 10}{1}-\frac{G \times 100}{1}=-110 G$
$=-110 \times 6.67 \times 10^{-11} \mathrm{J} \mathrm{kg}^{-1}$
$=-7.3 \times 10^{-9} \mathrm{J} \mathrm{kg}^{-1}$
Question 9
1 / -0

A particle moves along x-axis as x=4(t−2)+a(t−2)². Which of the following is true?

A
The initial velocity of the particle is 4 m s⁻¹

B
The acceleration of particle is 2a

C
The particle is at origin at t=0

D
None of the above

Solution

Given, x=4 (t−2)+a(t−2)²

v= $\frac{d x}{d t}$ =4+2a(t−2)

At t=0, v=4(1−a)

Acceleration a= $\frac{d^{2} x}{d t^{2}}$ =2a
Question 10
1 / -0

A heavy uniform chain lies on horizontal table top. If the coefficient of friction between the chain and the table surface is 0.25, then the maximum fraction of the length of the chain that can hang over one edge of the table is

A
20%

B
25%

C
35%

D
15%

Solution

The force of friction on the chain lying on the table should be equal to the weight of the hanging chain. Let

ρ= mass per unit length of the chain

μ= coefficient of friction

l= length of the total chain

x= length of hanging chain

Now, $\mu(l-x) \rho g=x \rho g$ or $\mu(l-x)=x$
$\Rightarrow \mu l=(\mu+1) x$ or $x=\frac{\mu l}{(\mu+1)}$
$\therefore x=\frac{0.25 l}{(0.25+1)}=\frac{0.25 l}{1.25}=0.2 l$
$\therefore \frac{x}{l}=0.2=20 \%$

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Physics Test - 18

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Physics Test - 18

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Question 1

2π mk2 r2

mk2 r2t

(mK4r2t5)5

Zero

Solution

Question 2

A to B and C to D

B to A and C to D

A to B and D to C

B to A and D to C

Solution

Question 3

R

2 R

3 R

4 R

Solution

Question 4

1

2

3

4

Solution

Question 5

25

32

35

23

Solution

Question 6

g cos θ

g sin θ

g cot θ

g tan θ

Solution

Question 7

1.6 cm

1.62 cm

1.622 cm

2 cm

Solution

Question 8

−7.3×10−7 J/k

−7.3×10−8 J/kg-7.3×10-8 J/kg

−7.3×10−9 J/k

−7.3×10−6 J/kg

Solution

Question 9

The initial velocity of the particle is 4 m s−1

The acceleration of particle is 2a

The particle is at origin at t=0

None of the above

Solution

Question 10

20%

25%

35%

15%

Solution

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2π mk² r²

mk² r²t

$\frac{(m K^{4} r^{2} t^{5})}{5}$

$\frac{2}{5}$

$\frac{3}{2}$

$\frac{3}{5}$

$\frac{2}{3}$

−7.3×10⁻⁷ J/k

−7.3×10⁻⁸ J/kg-7.3×10-8 J/kg

−7.3×10⁻⁹ J/k

−7.3×10⁻⁶ J/kg

The initial velocity of the particle is 4 m s⁻¹