Physics Test 190

Result

Physics Test 190

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Question 1
4 / -1

From the adjacent figure, the correct observation is:

A
Pressure at the bottom of the tank (a) is greater than at the bottom of the tank (b).

B
Pressure at the bottom of the tank (a) is lesser than at the bottom of (b).

C
Pressure depends upon the shape of the container.

D
Pressure at the bottom of (a) and (b) are the same.

Solution

Pressure = hρg i.e. pressure at the bottom is independent of the area of the bottom of the tank. It depends on the height upto which the tank is filled with water. As in both the tanks, the levels of water are the same, pressure at the bottom is also the same.
Question 2
4 / -1

A diver is 10 m below the surface of water. The approximate pressure experienced by the diver is-

A
10⁵ Pa

B
2 × 10⁵ Pa

C
3 × 10⁵ Pa

D
4 × 10⁵ Pa

Solution

P = P₀ + ρgh

Step 1: Put the values in above formula and find P.

= 10⁵+ 10³× 10 × 10

= 2 × 10⁵Pa
Question 3
4 / -1

The area of cross-section of the wider tube shown in the figure is 800 cm². If a mass of 12 kg is placed on the massless piston, then the difference in heights h of the levels of water in the two tubes will be:

A
10 cm

B
6 cm

C
15 cm

D
2 cm

Solution
Question 4
4 / -1

The height of a mercury barometer is 75 cm at sea level and 50 cm at the top of a hill. The ratio of density of mercury to that of air is 10⁴. The height of the hill is:

A
250 m

B
2.5 km

C
1.25 km

D
750 m

Solution
Question 5
4 / -1

A vertical U-tube of uniform inner cross section, contains mercury in both its arms. A glycerin (density = 1.3 g/ cm³) column of length 10 cm is introduced into one of its arms. Oil of density 0.8 gm/ cm³ is poured into the other arm until the upper surfaces of the oil and glycerin are at the same horizontal level. Find the length of the oil column. [Density of mercury = 13.6 g/cm³]

A
10.4 cm

B
8.2 cm

C
7.2 cm

D
9.6 cm

Solution

At the condition of equilibrium

Pressure at point A = Pressure at point B

P_A= P_B ⇒ 10 × 1.3 × g = h × 0.8 × g + (10 - h) × 13.6 × g

By solving we get h = 9.7 cm
Question 6
4 / -1

The value of the coefficient of volume expansion of glycerin is 5 x 10^-4 K^-1. The fractional change in the density of glycerin for a temperature increase of 40°C will be:

A
0.015

B
0.020

C
0.025

D
0.010

Solution
Question 7
4 / -1

When a uniform rod is heated, which of its following properties will increase as a result of it?

A
mass

B
weight

C
center of mass

D
moment of inertia

Solution

Step 1:

As the rod is heated its length increases.

Step 2: As the moment of inertia of the rod depends on its length so, it will changes.
Question 8
4 / -1

If two rods of length L and 2L having coefficients of linear expansion α and 2α respectively are connected so that their total length becomes 3L, the average coefficient of linear expansion of the composition of rods equals:

A

B

C

D
none of these

Solution
Question 9
4 / -1

Two rods, one made of aluminium and the other made of steel, having initial lengths l₁ and l₂ are connected together to form a single rod of length l₁ + l₂. The coefficient of linear expansion for aluminium and steel are α_a and α_s respectively. If the length of each rod increases by the same amount when their temperature is raised by t°C, then the ratio

A
α_s/α_a

B
α_a/α_s

C

D

Solution

Step 1: Find the change in the lengths of the two rods.
Question 10
4 / -1

A large steel wheel is to be fitted onto a shaft of the same material. At 27 °C, the outer diameter of the shaft is 8.70 cm and the wheel's central hole has a diameter of 8.69 cm. The shaft is cooled using ‘dry ice’. At what temperature of the shaft does the wheel slip on the shaft? (Assume the coefficient of linear expansion of the steel to be constant over the required temperature range and α steel = 1.20 × 10⁻⁵K⁻¹)

A
68°C

B
-70°C

C
-69°C

D
-67°C

Solution

The given temperature, T = 27° C = 27 + 273 = 300 K

Step 1: Use the formula of linear expansion.

We know that,

Δd = -d₁α_steel(T1 - T)

T₁ is temperature after cooling.

Step 2: Find the final temperature.

0.01 = -8.7 × 1.20 × 10^-5× (T₁ - 300)

(T₁ - 300) = -95.78

T₁ = 204.21 K

= -68.95°C

Therefore the wheel will slip on the shaft when the temperature is -69 °C.