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
A ball falls under gravity from a height of 10 m with an initial downward velocity u . It loses one third of its energy in collision and then rises back to 5 m . The initial velocity u is $\left(\mathrm{g}=10 \mathrm{~ms}^{-2}\right)$
$10 \mathrm{~ms}^{-1}$
$1 \mathrm{~ms}^{-1}$
$7.07 \mathrm{~ms}^{-1}$
$0.707 \mathrm{~ms}^{-1}$
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