Physics Test 163

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Physics Test 163

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Question 1
4 / -1

P is a point moving with constant speed 10 m/s such that its velocity vector always maintains an angle 60° with line OP as shown in figure (O is a fixed point in space). The initial distance between O and P is 100 m. After what time shall P reach O.

A
10 sec

B
15 sec

C
20 sec

D
20√3 sec

Solution
Question 2
4 / -1

A mosquito with 8 legs stands on water surface and each leg makes depression of radius ' a'. If the surface tension and angle of contact are ' T ' and zero respectively, then the weight of mosquito is :

A
8T. a

B
16 π T a

C

D

Solution

Given that the mosquito has 8 legs, each leg makes a depression of radius 'a' on the water surface. The angle of contact is given as zero, which means the water surface is flat at the point of contact.

Since the angle of contact is zero, the force exerted by surface tension on each leg is acting horizontally. Now, let's consider a single leg of the mosquito. The force exerted by surface tension on the leg (Fs) can be calculated as follows:

Fs = 2 * pi * a * T

Where,

Fs = Force exerted by surface tension on a single leg
a = radius of the depression
T = surface tension

Now, consider all 8 legs of the mosquito. The total force exerted by surface tension on all the legs is:

F_total = 8 * Fs

F_total = 8 * (2 * pi * a * T)

F_total = 16 * pi * a * T

Since the mosquito is at equilibrium, the total force exerted by surface tension must be equal to the weight of the mosquito (W).

W = F_total

W = 16 * pi * a * T

So, the weight of the mosquito is 16 * pi * a * T.
Question 3
4 / -1

The work done in increasing the size of a rectangular soap film with dimensions 8 cm × 3.75 cm to 10 cm × 6 cm is 2 × 10^–4 J. The surface tension of the film in N/m is :

A
1.65 × 10^–2

B
3.3 × 10^–2

C
6.6 × 10^–2

D
8.25 × 10^–2

Solution
Question 4
4 / -1

The property of surface tension is to________

A
increase the volume

B
decrease the volume

C
increase the surface area

D
decrease the surface area

Solution

Due to surface tension, the surface area tries to minimize itself.
Question 5
4 / -1

Two uniform rods of equal length but different masses are rigidly joined to form a L-shaped body, which is then pivoted about O as shown. If in equilibrium the body is in the shown configuration, ratio M/m will be:

A
2

B
3

C

D

Solution

In equilibrium, torques of forces mg and Mg about an axis passing through O balance each other.
Question 6
4 / -1

A large open tank has two small holes in its vertical wall as shown in figure. One is a square hole of side 'L' at a depth '4y' from the top and the other is a circular hole of radius 'R' at a depth ‘y’ from the top. When the tank is completely filled with water, the quantities of water flowing out per second from both holes are the same. Then, 'R' is equal to :

A

B
2πL

C

D
L/2π

Solution

Let v₁ and v₂ be the velocity of efflux from square and circular hole respectively.

S₁ and S₂ be cross-section areas of square and circular holes.

The volume of water coming out of square and circular hole per second is
Question 7
4 / -1

A stationary body explodes into two fragments of masses m₁ and m₂. If momentum of one fragment is p, the energy of explosion is :

A

B

C

D

Solution

By momentum conservation, the momentum of other fragment will be –p Total KE
Question 8
4 / -1

A shell of mass 2 m projected with a speed ' u ' at an angle θ to the horizontal explodes at the highest point of its motion into two pieces of mass ' m ' each. If one piece whose initial speed is zero, falls vertically, the distance at which the other piece will fall from the gun is given by :

A

B

C

D
None of these

Solution

The COM will fall at distance from the initial point, irrespective of explosion. Now one particle is at R/2 the other should be at 3R/2.
Question 9
4 / -1

A man of 80 kg attempts to jump from a small boat of mass 40 kg on to the shore. He can generate a relative velocity of 6 m/s between himself and boat. His velocity towards the shore is :

A
4 m/s

B
8 m/s

C
2 m/s

D
3 m/s

Solution

Let the velocity of boat be 'V' opposite to the man. By momentum conservation, –40V + 80 (6 – V) = 0

V = 4 m/s

velocity of the man = 6 – 4 = 2m/s
Question 10
4 / -1

A block of mass m slides along the track with coefficient of kinetic friction μ. A man pulls the block through a rope which makes an angle θ with the horizontal as shown in the figure. The block moves with constant speed V. Power delivered by the man is :

A
TV

B
TV cosθ

C
(T cos θ - μ mg) V

D
Zero

Solution