A vessel with water is placed on a weighing pan and it reads600 g. Now a ball of mass 40 g and density 0.80 g cm^-3 is sunk into the water with a pin of negligible volume, as shown in figure keeping it sunk. The weighting pan will show a reading

600 g
550 g
650 g
632 g

Solution

Volume of ball =⁴⁰⁄_0.8=50 cm³

Downthrust on water = 50 g.

Therefore reading is 650 g.

Water rises to a height of 10 cm in capillary tube and mercury falls to a depth of 3.1 cm in the same capillary tube. If the density of mercury is 13.6 and the angle of contact for mercury is 135°, the approximate ratio of surface tensions of water and mercury is

1 : 0.15
1 : 3
1 : 6
1.5 : 1

Solution

8 mercury drops coalesce to form one mercury drop, the energy changes by a factor of

Solution

A cylinder of height 20m is completely filled with water. The velocity of efflux of water (in ms–1) through a small hole on the side wall or the cylinder near its bottom is

10 m/s
20 m/s
25.5 m/s
5 m/s

Solution

Consider a 1 c.c. sample of air at absolute temperature T₀ at sea level and another 1 c.c. sample of air at a height where the pressure is one-third atmosphere. The absolute temperature T of the sample at the height is

equal to T₀/3
equal to 3/T₀
equal to T₀
cannot be determined in terms of T₀ from the above data

Solution

P α T

The level of water in a tank is 5m high.A hole of area 1 cm² is made in the bottom of the tank. The rate of leakage of water from the hole (g = 10 m/s²) is

10^-2 m³/s
10^-3 m³/s
10^-4 m³/s
10³ m³/s

Solution

The excess of pressure inside a soap bubble is twice the excess pressure inside a second soap bubble. The volume of the first bubble is n times the volume of the second where n is

0.125
0.250
1
2

Solution

The density of ice is x gram/litre and that of water is y gram/litre. What is the change in volume when m gram of ice melts?

mx y (x – y)
m/(y –x)
m(¹⁄_y-¹⁄_x)
(y – x)/x

Solution

Volume of m g of ice = m/x and volume of m g of water = m/y. So change in volume

= ^m⁄_y - ^m⁄_x=m(¹⁄_y-¹⁄_x)

An ice-berg floating partly immersed in sea water of density 1.03 g/cm³. The density of ice is 0.92 g/cm³. The fraction of the total volume of the iceberg above the level of sea water is

8.1%
11%
34%
0.8%

Solution

Let v be the volume of the ice-berg outside the seawater and V be the total volume of ice-berg. Then as per question

0.92 V= 1.03 (V – v) or v/V = 1 – 0.92/1.03

= 11/103

∴ (v/V) × 100 = 11 × 100 / 103 ≅ 11%

A beaker containing a liquid of density ρ moves up with an acceleration a. The pressure due to the liquid at a depth h below the free surface of the liquid is

h ρ g
h ρ (g – a)
h ρ (g + a)
\(2h\rho g\left ( \frac{g+a}{g-a} \right )\)

Solution

When a beaker containing a liquid of density ρ moves up with an acceleration a, it will work as a lift moving upward with acceleration a. The effective acceleration due to gravity in lift = (a + g)

∴ Pressure of liquid of height h = h ρ(a + g)

UnAttempted

0

Attempted

0

Correct

0

Wrong

0

Complete