VITEEE 2017 :: Physics :: General questions

• Physics - General questions

Two mercury drops (each of radius r) merge to form a bigger drop. The surface energy of the bigger drop, if T is the surface tension, is

Answer:

Explanation:

Let R be the radius of the bigger drop, then Volume of bigger drop = 2 x volume of small drop

$\frac{4}{3}\pi R^{3}=2\times\frac{4}{3}\pi r^{3}\Rightarrow R =2^{1/3} r$

Surface energy of bigger drop,

$E=4\pi R^{2}T = 4\times2^{2/3}\pi r^{2}T$ = $2^{8/3}\pi r^{2}T$

 A hollow insulated conduction sphere is given a positive charge of 10 μC. What will be the electric field at the centre of the sphere if its radius is 2 metres?

Answer:

Explanation:

Charge resides on the outer surface of a conducting hollow sphere of radius R. We consider a spherical surface of radius r < R. By Gauss theorem

$\int_{s}^{} \vec{E}.\vec{d}s = \frac{1}{\epsilon_{0}}$× Charge enclosed or

$E\times4\pi r^{2} =\frac{1}{\epsilon_{0}}\times0$

i.e electric field inside a hollow sphere is zero.

Light of wavelength 500 nm is incident on a metal with work function 2.28 eV. The de Broglie wavelength of the emitted electron is

Answer:

Explanation:

Given : work function Φ of metal = 2.28 eV

Wavelength of light λ = 500 nm = 500 x 10^-9m

$KE_{max} = \frac{hc}{\lambda}-\phi$

$KE_{max} = \frac{6.6\times10^{-34}\times3\times10^{8}}{5\times10^{-7}}-2.82$

$KE_{max} =2.48 - 2.28 = 0.2 eV$

$\lambda_{min} =\frac{h}{p}=\frac{h}{\sqrt{2m(KE)_{max}}}$

$\frac{\frac{20}{3}\times10^{-34}}{\sqrt{2\times9\times10^{-31}\times0.2\times1.6\times10^{-19}}}$

$\lambda_{min} =\frac{25}{9}\times10^{-9}$ = 2.80 x 10^-9nm

.'. λ ≥ 2.8 x 10^-9m

On a smooth plane surface (figure) two block A and B are accelerated up by applying a force 15 N on A. If mass of B is twice that of A, the force on B is

Answer:

Explanation:

The acceleration of both the blocks = $\frac{15}{3x}=\frac{5}{x}$

.'. Force on B = $\frac{5}{x}\times2x$ = 10N

The velocity of, efflux of a liquid through an orifice in the bottom of the tank does not depend upon

Answer:

Explanation:

v = velocity of efflux through an orifice = $\sqrt{2gH}$

It is independent of the size of the orifice.

• Physics - General questions