A nucleus with mass number 240 breaks into two fragments each of mass number 120, the binding energy per nucleon of unfragmented nucleus is 7.6 MeV while that of fragments is 8.5 MeV. The total gain in the binding energy in the process is:
(1) 0.9 MeV
(2) 9.4 MeV
(3) 804 MeV
(4) 216 MeV Continue reading A nucleus with mass number 240 breaks into ….
A spring is stretched by 5 cm by a force 10 N. The time period of the oscillations when a mass of 2 kg is suspended by it is:
(1) 0.0628 s
(2) 6.28 s
(3) 3.14 s
(4) 0.628 s Continue reading A spring is stretched by 5 cm by a force 10 N ….
An electromagnetic wave of wavelength $’\lambda ‘$ is incident on a photosensitive surface of negligible work function. If ‘m’ mass of photoelectron emitted from the surface has de-Broglie wavelength $\lambda_d$, then:
(1) $\lambda = \left( {\frac{{2m}}{{hc}}} \right){\lambda _d}^2$
(2) ${\lambda _d} = \left( {\frac{{2mc}}{h}} \right){\lambda ^2}$
(3) $\lambda = \left( {\frac{{2mc}}{h}} \right){\lambda _d}^2$
(4) $\lambda = \left( {\frac{{2h}}{{mc}}} \right){\lambda _d}^2$ Continue reading An electromagnetic wave of wavelength $’\lambda ‘$ ….
A cup of coffee cools from $90^{\circ}$C to $80^{\circ}$C in t minutes when the room temperature is $20^{\circ}$C. The time taken by a similar cup of coffee to cool from $80^{\circ}$C to $60^{\circ}$C at the same room temperature $20^{\circ}$C is:
(1) $\frac {13}{10}$t
(2) $\frac {13}{5}$t
(3) $\frac {10}{13}$t
(4) $\frac {5}{13}$t Continue reading A cup of coffee cools from $90^{\circ}$C ….
The number of photons per second on an average emitted by the source of monochromatic light of wavelength 600 nm, when it delivers the power of $3.3 \times 10^{-3}$ watt will be: ($h=6.6\times 10^{-34}$Js)
(1) $10^{18}$
(2) $10^{17}$
(3) $10^{16}$
(4) $10^{15}$ Continue reading The number of photons per second on an average ….
A particle moving in a circle of radius R with a uniform speed takes time T to complete one revolution. If this particle were projected with the same speed at an angle ‘$\theta $’ to the horizontal, the maximum height attained by it equals 4R. The angle of projection ‘$\theta $’ is then given by:
(1) $\theta = {\cos ^{ – 1}}\sqrt {\frac{{g{T^2}}}{{{\pi ^2}R}}} $
(2) $\theta = {\cos ^{ – 1}}\sqrt {\frac{{{\pi ^2}R}}{{g{T^2}}}} $
(3) $\theta = {\sin ^{ – 1}}\sqrt {\frac{{{\pi ^2}R}}{{g{T^2}}}} $
(4) $\theta = {\sin ^{ – 1}}\sqrt {\frac{{2g{T^2}}}{{{\pi ^2}R}}} $ Continue reading A particle moving in a circle of radius R ….
The resistance $R=\frac {V}{I}$, where $V=(50\pm 2) V$ and $I=(20 \pm 0.2 ) A $. The percentage error in R is ‘x’ %. The value of ‘x’ to the nearest integer is _ _ _ _ . Continue reading $R=\frac {V}{I}$
$V=(50\pm 2) V$
$I=(20 \pm 0.2 ) A $
% Error = ?
