A small hollow vessel which has a small circular hole of radius $r$ in its base, is immersed in a tank of oil of density $\rho$ and surface tension $T$. The oil will penetrate into the vessel at a depth of
$\frac{2 T}{r \rho g}$
$\frac{T}{r \rho g}$
$\frac{4 T}{r \rho g}$
$$ \frac{2}{\operatorname{r} \rho g T} $$
Two-point charges of equal magnitude 0.01 C and opposite in sign are separated by 0.2 mm , forming an electric dipole. The electric dipole moment of the dipole is:
$2 \times 10^{-3} \mathrm{Cm}$ along the line joining the two charges and the direction of the dipole moment vector is from negative to positive 0.01 C
$4 \times 10^{-6} \mathrm{Cm}$ along the line joining the two charges and the direction of the dipole moment vector is from positive to negative 0.01 C
$2 \times 10^{-6} \mathrm{Cm}$ along the line joining the two charges and the direction of the dipole moment vector is from negative to positive 0.01 C
$2 \times 10^6 \mathrm{Cm}$ along the line joining the two charges and the direction of the dipole moment vector is from positive to negative 0.01 C
1 kg of water at $100^{\circ} \mathrm{C}$ is converted to steam at the same temperature. Volume of 1 cc of water changes to $1671 \times 10^3 \mathrm{cc}$ on boiling. The change in internal energy of the system is (Latent Heat of vaporisation of water is $22.68 \times 10^5 \mathrm{~J} \mathrm{~kg}^{-1} ; 1 \mathrm{~atm}=1.0 \times 10^5 \mathrm{~Pa}$ )
$21.01 \times 10^3 J$
$19.01 \times 10^5 J$
$20.1 \times 10^5 J$
$21.01 \times 10^5 J$
A mercury-198 nucleus is bombarded by a neutron, which causes a nuclear reaction
$$ n_0^1+\mathrm{Hg}_{80}^{198} \longrightarrow A u_{79}^{197}+X $$
What is the unknown product particle $X$ ?
Alpha particle
Beta particle
Deuteron
Proton
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