Physics Test 122

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Physics Test 122

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Question 1
4 / -1

The volumes of containers A and B, connected by a tube and a closed valve are V and 4 V, respectively. Both the containers A and B have the same ideal gas at pressures (temperatures) 5.0 ×10⁵ Pa(300 K) and 1.0 ×10⁵Pa (400 K), respectively. The valve is opened to allow the pressure to equalise, but the temperature of each container is kept constant at its initial value. Find the common pressure in the containers.

A
2.5 × 10⁵Pa

B
2.0 × 10⁵Pa

C
3.0 × 10⁵ Pa

D
1.5 × 10⁵Pa

Solution

∴ p₁ > p₂ so, the equalise pressure of the p is reduced from p₁
Question 2
4 / -1

A block of mass 10 kg is suspended through two light spring balances as shown in figure

A
Both the scales will read 10 kg.

B
Both the scales will read 5 kg.

C
The upper scale will read 10 kg and the lower zero.

D
The readings may be anything but their sum will be 10 kg.

Solution

The FED of the spring balances and the block are as shown in figure.

T₁=10g T₁=T₂ ⇒ T₂=10g where, T₁ and T₂are readings of spring balances as shown in figure.
Question 3
4 / -1

If E = 100 sin (100t) volt and are the instantaneous values of voltage and current, then the R.M.S values of voltage and current are respectively.

A
70.7 V, 70.7 mA

B
70.7 V, 70.7A

C
141.4 V, 141.4 mA

D
141.4 V, 141.4 A

Solution

The instantanous valur of voltae is

E= 100sin (100 t) V ...(1)

(compare it with

E = E₀sin (ωt) V

we get,

E₀= 100 V, ω = 100 rads^-1

the r.m.s value of voltage is

compare it with, I=I₀sin(ω t + Φ)

we get

I₀ = 100 mA, ω = 100 rads^-1

The r.m.s value of current is
Question 4
4 / -1

Two spherical bodies of mass M and 5 M and radii R and 2 R, respectively are released in free space with initial separation between their centres equal to 12 R. If they attract each other due to gravitational force only, then the distance covered by the smaller body just before collision is

A
2.5 R

B
4.5 R

C
7.5 R

D
1.5 R

Solution

Initial separation between both centers = 12R
Final separation (when collision occurs) = 3R
Thus both the centers need to travel 9R combinely in such a way that their COM does not move as no external force is acting over the 2 mass system.
Thus let say if smaller mass travel x distance, bigger would eventually travel rest 9R - x
Thus we we write the equation of displacement of the COM, and taking direction of displacement of bigger mass as positive, we get
M (-x) + 5M (9R - x) / (M + 5M) = 0
Thus we get, -Mx + 45MR - 5Mx = 0
We get 6x = 45R
Thus x = 7.5R
Question 5
4 / -1

Mark the correct option.

A
Behaviour of moving charge along the circle is equivalent to current carrying circular coil for instantaneous value as well as for average value of magnetic field on its axis

B
If two identical charge particles are moving symmetrically along a circle then its behaviour with respect to magnetic field on the axis is equivalent to current carrying coil for both average and instantaneous value

C
For instantaneous value of magnetic field on the axis, the charge moving along a circle is equivalent to current carrying circular coil

D
Both b and c

Solution

A charge moving along a circle is equivalent to a current carrying coil, but with respect to magnetic field on the axis of circle. It is equivalent only for the average value of the magnetic field and not for instantaneous values. While if two charge particles are moving symmetrically along a circle at diametrically opposite points then the average as well as instantaneous magnetic field on its axis is same as due to a current carrying coil.
Question 6
4 / -1

The length of a given cylindrical wire is increased by 100%. Due to the consequent decrease in diameter the change in the resistance of the wire will be

A
200%

B
100%

C
50%

D
300%

Solution
Question 7
4 / -1

Students I, II and III perform an experiment for measuring the acceleration due to gravity (g) using a simple pendulum. They use different lengths of the pendulum and /or record time for different number of oscillations. The observations are shown in the table.

Least count for length = 0.1 cm

Least count for time = 0.1 s

If E_I, E_II and E_IIIare the percentage errors in g, i.e.,

for students I, II and III, respectively, then

A
E_I = 0

B
E_I is minimum

C
E_I= E_II

D
E_IIis maximum

Solution

The period of oscillation (T) of a simple pendulum of length ℓ is given by

T = 2π√(ℓ/g)

Therefore, g = 4π2 ℓ/T2 so that the fractional error in g is given by

∆g/g = (∆ℓ/ℓ) + 2(∆T/T)

[The above expression is obtained by taking logarithm of both sides and then differentiating. Note that the sign of the second term on the RHS is changed from negative to positive since we have to consider the maximum possible error].

Here ∆ℓ = 0.1 cm and ∆T = 0.1 s

The percentage error is 100 times the fractional error so that

E_I = ∆g/g = [(0.1/64) + 2(0.1/128)]×100 = 5/16 %,

E_II = ∆g/g = [(0.1/64) + 2(0.1/64)]×100 = 15/32 % and

E_III = ∆g/g = [(0.1/20) + 2(0.1/36)]×100 = 19/18 %

Thus E_I is minimum.
Question 8
4 / -1

Write down the expression for capacitance of a spherical capacitor whose conductors radii are R₁and R₂(R₂>R1),when inner sphere is grounded.

A

B

C

D

Solution
Question 9
4 / -1

A mass of M kg is suspended by a weightless string. The horizontal force that is required to displace it until the string makes an angle of 45° with the initial vertical direction is

A

B
Mg √2

C
Mg /√2

D

Solution

Here, position 1 and position 2 can be calculated by using

work- energy Theorem
Question 10
4 / -1

Acceleration of each block is given as g/5√2. Find the magnitude and direction of force exerted by string on pulley. (μ = 0.4)

A
6 mg/5, downward direction.

B
6 mg/5, upward direction.

C
6 mg/5, horizontal direction.

D
Not possible to have acceleration.

Solution

Let coefficient of friction be u. and and 3m block is moving down the incline, then Acceleration

Force exerted by string/on pulley is √2T as shown in figure

∴ F = 6mg/5