A capacitor of capacitance \(C=1\, \mu \,{F}\) is suddenly connected to a battery of \(100\, volt\) through a resistance \({R}=100\, \Omega\). The time taken for the capacitor to be charged to get \(50 \,{V}\) is \(....\,\times \,10^{-4}\,s.\) [Take \(\ln 2=0.69]\)

In an adiabatic process, the density of a diatomic gas becomes \(32\) times its initial value. The final pressure of the gas is found to be \(n\) times the initial pressure. The value of \(n\) is

The approximate height from the surface of earth at which the weight of the body becomes \(\frac{1}{3}\) of its weight on the surface of earth is \(..........km\) : [Radius of earth \(R =6400 \,km\) and \(\sqrt{3}=1.732]\)

Consider two charged metallic spheres \(S_{1}\) and \(\mathrm{S}_{2}\) of radii \(\mathrm{R}_{1}\) and \(\mathrm{R}_{2},\) respectively. The electric \(\left.\text { fields }\left.\mathrm{E}_{1} \text { (on } \mathrm{S}_{1}\right) \text { and } \mathrm{E}_{2} \text { (on } \mathrm{S}_{2}\right)\) on their surfaces are such that \(\mathrm{E}_{1} / \mathrm{E}_{2}=\mathrm{R}_{1} / \mathrm{R}_{2} .\) Then the ratio \(\left.\mathrm{V}_{1}\left(\mathrm{on}\; \mathrm{S}_{1}\right) / \mathrm{V}_{2} \text { (on } \mathrm{S}_{2}\right)\) of the electrostatic potentials on each sphere is 

A resonance tube is old and has jagged end. It is still used in the laboratory to determine velocity of sound in air. A tuning fork of frequency \(512\,Hz\) produces first resonance when the tube is filled with water to a mark \(11\,cm\) below a reference mark, near the open end of the tube. The experiment is repeated with another fork of frequency \(256\,Hz\) which produces first resonance when water reaches a mark \(27\,cm\) below the reference mark. The velocity of sound in air, obtained in the experiment, is close to .... \(ms^{-1}\)

A metallic sphere cools from \(50^{\circ} C\) to \(40^{\circ} C\) in \(300 \,s.\) If atmospheric temperature around is \(20^{\circ} C ,\) then the sphere's temperature after the next \(5\) minutes will be close to\(.....C\)

Two charges \(7 \mu \mathrm{c}\) and \(-4 \mu \mathrm{c}\) are placed at \((-7 \mathrm{~cm}, 0,0)\) and \((7 \mathrm{~cm}, 0,0)\) respectively. Given, \(\epsilon_0=8.85 \times 10^{-12} \mathrm{C}^2 \mathrm{~N}^{-1} \mathrm{~m}^{-2}\), the electrostatic potential energy of the charge configuration is :

In the following circuit, the correct relation between output \(( Y )\) and inputs \(A\) and \(B\) will be

For a given transistor amplifier circuit in \(CE\) configuration \(V_{C C}=1 V , R_c=1 k \Omega, R_b=100 k \Omega\) and \(\beta=100\). Value of base current \(I_b\) is