Physics Test 189

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Physics Test 189

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Weekly Quiz Competition

Question 1
4 / -1

A U tube with both ends open to the atmosphere, is partially filled with water. Oil, which is immiscible with water, is poured into one side until it stands at a distance of 10 mm above the water level on the other side. Meanwhile the water rises by 65 mm from its original level (see diagram). The density of the oil is

A
425 kgm^-3

B
800 kgm⁻³

C
928 kgm⁻³

D
650 kgm⁻³

Solution
Question 2
4 / -1

A rectangular film of liquid is extended from (4cm × 2cm) to (5cm × 4cm). If the work done is 3 × 10⁻⁴J, the value of the surface tension of the liquid is

A
0.250Nm⁻¹

B
0.125NM⁻¹

C
0.2Nm⁻¹

D
8.0Nm⁻¹

Solution

Work done = Surface tension of film x Change in area of the film
Question 3
4 / -1

Two identical bodies are made of a material for which the heat capacity increases with temperature. One of these is at 100°C, while the other one is at 0°C. If the two bodies are brought into contact, then, assuming no heat loss, the final common temperature is

A
50°C

B
More than 50°C

C
less than 50°C but greater than 0°C

D
0°C

Solution

Since, heat capacity of material increases with increase in temperature so, body at 100 °C has more heat capacity than body at 0°C. Hence, final common temperature of the system will be closer to 100\°C

∴ T_c > 50°C
Question 4
4 / -1

Coefficient of linear expansion of brass and steel rods are α₁ and α₂. Length of brass and steel rods are l₁ and l₂ respectively. If (l₂ − l₁) is maintained same at all temperatures, which one of the following relations holds good?

A
α₁²l₂ = α₂²l₁

B
α₁l₁ = α₂l₂

C
α₁l₂ = α₂l₁

D
α₁l₂² = α₂l₁²

Solution

Linear expansion of brass = α₁

Linear expansion of steel = α²

Length of brass rod = l₁

Length of steel rod = l₂

On increasing the temperature of the rods by ΔT, new lengths would be

l₁′ = l₁(1 + α₁ΔT) ........(i)

l₂′ = l₂(1 + α₂ΔT) .........(ii)

Subtracting eqn. (i) from eqn. (ii), we get

l₂′ − l₁′ = (l₂ − l₁) + (l₂α₂ − l₁α₁)ΔT

According to question,

l₂′ − l₁′ = l₂ − l₁ (for all temperatures)

∴ l₂α₂− l₁α₁ or l₁α₁ = l₁α₂
Question 5
4 / -1

A black body is at a temperature of 5760 K. The energy of radiation emitted by the body at wavelength 250nm is U₁, at wavelength 500 nm is U₂ and that at 1000 nm is U₃ Wien's constant, b = 2.88 × 10⁶nmK. Which of the following is correct?

A
U₁ > U₂

B
U₂ > U₁

C
U₁ = 0

D
U₃ = 0

Solution
Question 6
4 / -1

An airplane of a total wing area of 120 m² is in a level flight at some height. If the pressure difference between the upper and lower surface is 2.5 kPa, then the mass of the airplane is: [Take g = 10 ms-2]

A
2 x 10³ kg

B
3 x 10⁴ kg

C
5 x10⁴ kg

D
7 x 10⁷ kg

Solution
Question 7
4 / -1

A liquid is poured into three vessels of the same base area and equal heights as shown in the figure, then:

A
Maximum force on the base will be for vessel C

B
Maximum force on the base will be for vessel B

C
Maximum force on the base will be for vessel A

D
Force on the base will be equal for all the vessels

Solution

Step 1: Find the pressure at the bottom in all cases

P_b = P₀ + ρgh

Step 2: Find thrust at the base of each vessel

F = P_bA

= P_bA

As the pressure and the area is the same for all the cases. So, the force on the base will be equal for all the vessels.
Question 8
4 / -1

If pressure at half the depth of a lake is equal to 2/3rd the pressure at the bottom of the lake, then the depth of the lake is:

A
10m

B
20m

C
60m

D
30m

Solution
Question 9
4 / -1

The value of g at a place decreases by 2%. Then, the barometric height of mercury:

A
increases by 2%.

B
decreases by 2%.

C
remains unchanged.

D
sometimes increases and sometimes decreases.

Solution

If the value of g is decreased by 2% then h will increase by 2%.
Question 10
4 / -1

A siphon in use is demonstrated in the following figure. The density of the liquid flowing in the siphon is 1.5 gm/cc. The pressure difference between the point P and S will be:

A
10⁵ N / m

B
2 × 10⁵ N / m

C
Zero

D
Infinity

Solution

As both the points are at the surface of the liquid and in the open atmosphere. So, both the points possess similar pressure equal to 1 atm. Hence the pressure difference will be zero.