SECTION - B [PHYSICS - NUMERIC] - JEE STD 11 Science Questions

Question 14 Marks

The ratio of the power of a light source $S_1$ to that the light source $S _2$ is $2 . S _1$ is emitting $2 \times 10^{15}$ photons per second at 600 nm . If the wavelength of the source $S _2$ is 300 nm , then the number of photons per second emitted by $S _2$ is ___________ $\times 10^{14}$. (5)

Answer

(5)
Sol. Since power emitting by a source is given as
$\begin{array}{l}
=\frac{\text { Total energy emitted }}{\text { time }} \\
=\frac{\left(E_1 \text { photon }\right) \times \text { Number of photons }(N)}{t} \\
P_1=\left(E_1\right) n
\end{array}
$
$
\begin{array}{l}
\frac{P_1}{P_2}=\frac{\left(E_1\right) n_1}{\left(E_2\right) n_2}=\frac{\left(\frac{hC}{\lambda_1}\right) n_1}{\left(\frac{hC}{\lambda_2}\right) n_2} \\
\frac{P_1}{P_2}=\left(\frac{\lambda_2}{\lambda_1}\right) \frac{n_1}{n_2}
\end{array}
$
Substituting the given values$
\begin{array}{l}
2=\left(\frac{300}{600}\right) \times \frac{2 \times 10^{15}}{n_2} \\
n_2=\frac{1}{2} \times 10^{15}=5 \times 10^{14} \text { Photon } / sec
\end{array}
$

View full question & answer→

Question 24 Marks

The increase in pressure required to decrease the volume of a water sample by $0.2 \%$ is $P \times 10^{\circ} Nm ^{-2}$. Bulk modulus of water is $2.15 \times 10^9 Nm ^{-2}$. The value of $P$ is _____________

Answer

(43)
Sol. Since bulk modulus is given as
$
\begin{array}{l}
B=\frac{-\Delta P}{\left(\frac{\Delta V}{V}\right)} \\
2.15 \times 10^9=\frac{-\Delta P}{-\left(\frac{0.2}{100}\right)} \\
\Delta P=2.15 \times 10^9 \times 2 \times 10^{-3} \\
=4.3 \times 10^6=43 \times 10^5 N / m^2
\end{array}
$

View full question & answer→

Question 34 Marks

A tightly wound long solenoid carries a current of
1.5 A. An electron is executing uniform circular motion inside the solenoid with a time period of 75 ns . The number of turns per metre in the solenoid is __________ .
[Take mass of electron $m _e=9 \times 10^{31} kg$, charge of electron $\left| q _e\right|=1.6 \times 10^{-19} C$,$
\left.\mu_{o}=4 \pi \times 10^{-7} \frac{N}{A^2}, 1 ns=10^{-9} s\right]
$

Answer

(250)
Sol. Since time period of a revolving charge is $\frac{2 \pi m}{ qB }$
Where $B =$ magnetic field
$
\begin{array}{l}
\text { due to a solenoid }=\mu_0 nI \\
\therefore T=\frac{2 \pi m}{q\left(\mu_0 nI\right)} \\
75 \times 10^{-9}=\frac{(2 \pi)\left(9 \times 10^{-31}\right)}{1.6 \times 10^{-19} \times 4 \pi \times 10^{-7} \times n \times 1.5} \\
N=250
\end{array}
$

View full question & answer→

Question 44 Marks

A string of length $L$ is fixed at one end and carries a mass of M at the other end. The mass makes $\left(\frac{3}{\pi}\right)$ rotations per second about the vertical axis passing through end of the string as shown. The tension in the string is ___________ ML.

Answer

(36)
Sol.

$T \cos \theta= mg\qquad\ldots(1) $
$T \sin \theta= M \omega^2 R\qquad\ldots(2) $
Using equation (2)
$T \sin \theta= M \omega^2(L \sin \theta)$
$T = M \omega^2 L= M \left(\frac{3}{\pi} \times 2 \pi\right)^2 L$
$
T=36 ML
$

View full question & answer→

Question 54 Marks

Acceleration due to gravity on the surface of earth is ' g '. If the diameter of earth is reduced to one third of its original value and mass remains unchanged, then the acceleration due to gravity on the surface of the earth is ____________ g.

Answer

(9)
Sol. $\because$ acceleration due to gravity on surface is given by
$
g=\frac{GM}{R_{e}^2}
$
Now since diameter is reduced to $1 / 3^{\text {nd }}$, radius also reduces to $1 / 3^{\text {nd }}$, keeping mass constant
New value of acceleration due to gravity on Earth's surface is
$g^{\prime}=\frac{GM}{\left(\frac{R_{e}}{3}\right)^2}=9 \frac{GMe}{R_{c}^2}=9 g
$

View full question & answer→

JEE SECTION - B [PHYSICS - NUMERIC]

SECTION - B [PHYSICS - NUMERIC]

Take a timed test