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

Question 14 Marks

The temperature of 1 mole of an ideal monoatomic gas is increased by $50^{\circ} \mathrm{C}$ at constant pressure. The total heat added and change in internal energy are $E_{1}$ and $E_{2}$, respectively. If $\frac{E_{1}}{E_{2}}=\frac{x}{9}$ then the value of $x$ is ____________ .

Answer

(15)
Sol. Given that process is isobaric $\Delta \mathrm{T}=50^{\circ} \mathrm{C}$
Q in isobaric process $=\mathrm{nC}_{\mathrm{p}} \Delta \mathrm{T}=\mathrm{E}_{1}$
$\Delta \mathrm{U}$ in isobaric process $=\mathrm{nC}_{\mathrm{v}} \Delta \mathrm{T}=\mathrm{E}_{2}$
$\therefore \frac{E_{1}}{E_{2}}=\frac{C_{P}}{C_{V}}=\gamma$
Given, gas is monoatomic
$
\begin{aligned}
\therefore \gamma & =1+\frac{2}{\mathrm{f}} \\
& =1+\frac{2}{3} \\
& =\frac{5}{3}
\end{aligned}
$
Now, as per question.
$\frac{5}{3}=\frac{x}{9}$
$\mathrm{x}=15$

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Question 24 Marks

A current of 5 A exists in a square loop of side $\frac{1}{\sqrt{2}} \mathrm{~m}$. Then the magnitude of the magnetic field $B$ at the centre of the square loop will be $p \times 10^{-6} \mathrm{~T}$. where, value of $p$ is ____________ .
[Take $\mu_{0}=4 \pi \times 10^{-7} \mathrm{~T} \mathrm{~mA}^{-1}$ ].

Answer

(8)
Sol.

Let $B$ be the magnetic field due to single side then
$B=\frac{\mu_{0} \mathrm{i}}{4 \pi \mathrm{~d}}\left(\sin \theta_{1}+\sin \theta_{2}\right)$
$=\frac{10^{-7} \times 5 \times 2}{\frac{1}{2 \sqrt{2}}} \times \frac{1}{\sqrt{2}}=2 \times 10^{-6}$
$\therefore \mathrm{B}_{\text {net }}$ at centre $\mathrm{O}=4 \mathrm{~B}$
$=8 \times 10^{-6}$
$\therefore \mathrm{P}=8$ Ans.

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Question 34 Marks

A wire of resistance 9 is bent to form an equilateral triangle. Then the equivalent resistance across any two vertices will be ____________ ohm.

Answer

(2)
Sol.

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Question 44 Marks

The least count of a screw guage is 0.01 mm . If the pitch is increased by $75 \%$ and number of divisions on the circular scale is reduced by $50 \%$, the new least count will be ____________ $\times 10^{-3} \mathrm{~mm}$.

Answer

(35)
Sol. Given least count of Screw Gauge $=0.01 \mathrm{~mm}$
L.C $=\frac{(\text { pitch })}{\text { No. of circular turn }}=\frac{\mathrm{P}}{\mathrm{N}}=0.01 \mathrm{~mm}$
New pitch $=\frac{P(1+0.75)}{N(1-0.5)}=\frac{P}{N}\left[\frac{1.75}{0.5}\right]$
$=(0.01) 3.5$
$=0.035 \mathrm{~mm}$
$=35 \times 10^{-3} \mathrm{~mm}$
$\therefore$ Ans. is 35

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Question 54 Marks

A square loop of sides $\mathrm{a}=1 \mathrm{~m}$ is held normally in front of a point charge $\mathrm{q}=1 \mathrm{C}$. The flux of the electric field through the shaded region is $\frac{5}{\mathrm{p}} \times \frac{1}{\varepsilon_{0}} \frac{\mathrm{Nm}^{2}}{\mathrm{C}}$, where the value of p is ___________ .

Answer

(48).
Sol.

Total flux through square $=\frac{\mathrm{q}}{\in_{0}}\left(\frac{1}{6}\right)$
Lets divide square is 8 equal parts.
Flux is same for each part.
$\therefore$ Flux through shaded portion is $\frac{5}{8}$ (Total flux)
$
=\frac{5}{8} \times \frac{\mathrm{q}}{\in_{0}} \frac{1}{6}=\frac{5}{48} \frac{1}{\in_{0}}
$
$\therefore$ required Ans. is 48

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JEE SECTION - B [PHYSICS - NUMERIC]

SECTION - B [PHYSICS - NUMERIC]

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