## A bakelite beaker has volume capacity of 500 cc at 30 °C .When it is partially filled with volume Vm at 30°C of mercury ….

Q: A bakelite beaker has volume capacity of 500 cc at 30 °C .When it is partially filled with volume Vm at 30°C of mercury . , it is found that the unfilled volume of the beaker remains constant as the temperature is varied . If γbeaker = 6 × 10-6/°C and γmercury = 1.5 × 10-4/°C , where γ is the coefficient of volume expansion , then Vm in cc is close to ……

Click to See Solution :
Ans: (20)

Sol: Given V = 500 cc , T = 30 °C

γbeaker = 6 × 10-6/°C

and γmercury = 1.5 × 10-4/°C

V – Vm = V’ – Vm

Now , V’ = V(1 + γb ΔT)

and , Vm‘ = Vm(1 + γm ΔT)

V – Vm = V + V γb ΔT – Vm – Vm γm ΔT)

$\displaystyle V_m = V \frac{\gamma_b}{\gamma_m}$

$\displaystyle V_m = 500 \times \frac{6 \times 10^{-6}}{1.5 \times 10^{-4}}$

Vm = 20 cc

## A non-isotropic solid metal cube has coefficients of linear expansion as 5 × 10-5 °/C  along the x -axis ….

Q: A non-isotropic solid metal cube has coefficients of linear expansion as 5 × 10-5 °/C  along the x -axis and 5 × 10-6 °/C  along the y -axis and z -axis . If the coefficient of volume expansion of solid is C × 10-5 °/C then the value of C is ….

Click to See Solution :
Ans: (60)

Sol: αlx = 5 × 10-5 °/C

αly = αlz = 5 × 10-6 °/C

αv = αlx + αly + αlz

αv = αlx + 2 αly

αv = 5 × 10-5 + 2 × 5 × 10-6

= 6 × 10-5 °/C

= 60 × 10-6 °/C

C = 60

## A bullet of mass 5 g , travelling with a speed of 210 m/s , strikes a fixed wooden target …..

Q: A bullet of mass 5 g , travelling with a speed of 210 m/s , strikes a fixed wooden target . One half of its kinetic energy is converted into heat in the bullet while the other half is converted into heat in the wood . The rise of temperature of the bullet if the specific heat of its material is 0.030 cal/g°C . (1 cal = 4.2 × 107 ergs)

(a) 87.5 °C

(b) 83.3 °C

(c) 119.2 °C

(d) 38.4 °C

Click to See Solution :
Ans: (a)
Sol: Given m = 5 g , v = 210 m/s

As W = J Q

$\displaystyle \frac{1}{2}m v^2 = J \times m \times s \times \Delta T$

$\displaystyle \Delta T = \frac{v^2}{4 J s }$

$\displaystyle \Delta T = \frac{(210)^2}{4 J s}$

= 87.5 °C

## A leakproof cylinder of length 1 m , made of a metal which has very low coefficient of expansion is floating vertically ….

Q: A leakproof cylinder of length 1 m , made of a metal which has very low coefficient of expansion is floating vertically in water at 0°C such that its height above the water surface is 20 cm . When the temperature of water is increased to 4°C , the height of the cylinder above the water surface becomes 21 cm . The density of water at T = 4°C , relative to density at T = 0°C is close to

(a) 1.03

(b) 1.01

(c) 1.26

(d) 1.04

Click to See Solution :
Ans: (b)

Sol: Let A be the area of cross section of cylinder

As height of cylinder above water surface = 20 cm

height of cylinder inside the water surface = 80 cm

When water at 0°C ,

m g = 80 × A × ρ0°C × g …(i)

When water at 4°C ,

m g = 79 × A × ρ4°C × g …(ii)

From (i) & (ii)

$\displaystyle \frac{\rho_{4^oC}}{\rho_{0^oC}} = \frac{80}{79} = 1.01$

## Heat is being supplied at a constant rate to a sphere of ice which is melting at the rate of 0.1 g/sec . It melts completely in 100 sec …

Q: Heat is being supplied at a constant rate to a sphere of ice which is melting at the rate of 0.1 g/sec . It melts completely in 100 sec . The rate of rise of temperature thereafter will be (Assume no loss of heat)

(a) 0.8 °C/sec

(b) 5.4 °C/sec

(c) 3.6 °C/sec

(d) None

Ans: (a)

Sol: m/t = 0.1

m = 0.1 t = 0.1 × 100 = 10 gm

Q = mL

$\displaystyle \frac{dQ}{dt} = L\frac{dm}{dt}$

$\displaystyle \frac{dQ}{dt} = 80 \times 0.1$

= 8 cal/sec

Again , $\displaystyle \frac{dQ}{dt} = m s \frac{d\theta}{dt}$

$\displaystyle \frac{d\theta}{dt} = \frac{\frac{dQ}{dt}}{m s}$

$\displaystyle \frac{d\theta}{dt} = \frac{8}{10 \times 1}$

= 0.8