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

Question 14 Marks

An inductor of self inductance 1 H connected in series with a resistor of $100 \pi$ ohm and an ac supply of $100 \pi$ volt, 50 Hz. Maximum current flowing in the circuit is ____________ A.

Answer

1
Impedance of circuit
$Z=\sqrt{R^2+\left(X_{L}\right)^2}=\sqrt{R^2+\left(\omega_{L}\right)^2} $
$=\sqrt{(100 \pi)^2+(2 \pi \times 50 \times 1)^2} $
$=\sqrt{(100 \pi)^2+(100 \pi)^2} $
$=\sqrt{2} \times 100 \pi $
$I_{rms}=\frac{V}{2}=\frac{100 \pi}{\sqrt{2} \times 100 \pi}=\frac{1}{\sqrt{2}} $
$I_{\max }=\sqrt{2} I_{rms}=\sqrt{2} \times \frac{1}{\sqrt{2}}=1 \text { Ampere }$

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Question 24 Marks

In a Young's double slit experiment, two slits are located 1.5 mm apart. The distance of screen from slits is 2 m and the wavelength of the source is 400 nm. If the 20 maxima of the double slit pattern are contained within the centre maximum of the single slit diffraction pattern, then the width of each slit is $x \times 10^{-3} cm$, where x-value is ___________ .

Answer

15
Width of 20 maxima of double slit $=$ width of central maxima of single slit
$
\begin{array}{l}
\frac{20 \lambda D}{d}=\frac{2 \lambda D}{a} \\
\frac{10}{d}=\frac{1}{a} \\
a=\frac{d}{10}=\frac{1.5 \times 10^{-1}}{10} cm=15 \times 10^{-3} cm
\end{array}
$
Value of $x$ is 15

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Question 34 Marks

If an optical medium possesses a relative permeability of $\frac{10}{\pi}$ and relative permittivity of $\frac{1}{0.0885}$, then the velocity of light is greater in vacuum than that in this medium by ____________ times.
$
\begin{array}{l}
\left(\mu_0=4 \pi \times 10^{-7} H / m, \quad \in_0=8.85 \times 10^{-12} F / m\right. \quad \left.c=3 \times 10^8 m / s\right)
\end{array}
$

Answer

6
Since velocity of light in terms of $\mu$ $\&$ $E$ is
$V=\frac{1}{\sqrt{\mu \in}}=\frac{1}{\sqrt{\mu_0 \mu_{ r }}} \times \frac{1}{\sqrt{\in_0 \in_{ r }}}$
$=\frac{1}{\sqrt{\mu_r \in_r}} \times \frac{1}{\sqrt{\mu_0 \in_0}}$
$=\frac{C}{\sqrt{\mu_r \in_r}}=\frac{C}{\sqrt{\frac{10}{\pi} \times \frac{1}{0.0885}}}$
$=\frac{C}{\sqrt{36}}=\frac{C}{6}$
$
\begin{aligned}
V & =\frac{C}{6} \\
C & =6 V
\end{aligned}
$
Velocity of light in vacuum is greater by 6 times the velocity of light in medium

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Question 44 Marks

A solid sphere with uniform density and radius R is rotating initially with constant angular velocity $\left(\omega_1\right)$ about its diameter. After some time during the rotation its starts loosing mass at a uniform rate, with no change in its shape. The angular velocity of the sphere when its radius become $R / 2$ is $x \omega_1$. The value of $x$ is __________ .

Answer

32
When sphere is of radius R , its mass is M , when radius is reduced to $\frac{R}{2}$, mass will reduced to $\frac{M}{8}$
Now by conservation of angular momentum $\left(\tau_{\text {ext }}=0\right)$
$L _1= L _2$
$I _1 \omega_1= I _2 \omega_2$
$\left(\frac{2}{5} MR ^2\right) \omega_1=\left(\frac{2}{5}\left(\frac{ M }{8}\right)\left(\frac{ R }{2}\right)^2\right) \omega_2$
$\omega_2=32 \omega_1$ value of $x$ is 32

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Question 54 Marks

A particle of charge $1.6$ $\mu C$ and mass $16$ $\mu g$ is present in a strong magnetic field of 6.28 T. The particle is then fired perpendicular to magnetic field. The time required for the particle to return to original location for the first time is __________ s.$(\pi=3.14)$

Answer

Angle between $\overrightarrow{ V }$ of charge & $\overrightarrow{ B }$ is $90^{\circ}$ motion will be uniform circular motion time period is given by
$T =\frac{2 \pi m}{ qB }=\frac{2 \pi \times 16 \times 10^{-9} kg}{1.6 \times 10^{-6} \times 6.28}$
$T =0.01$ seconds
NTA Answer is 10

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JEE SECTION - B [PHYSICS - NUMERIC]

SECTION - B [PHYSICS - NUMERIC]

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