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JEE SECTION - B [PHYSICS - NUMERIC]

JEE Main 7-April-2025 Paper - Shift 1 · Gujarat Board (GSEB) STD 11 Science (GSEB - English Medium). 5 questions.

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SECTION - B [PHYSICS - NUMERIC]

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Question 24 Marks
A wire of length 10 cm and diameter 0.5 mm is used in a bulb. The temperature of the wire is $1727^{\circ} \mathrm{C}$ and power radiated by the wire is 94.2 W. Its emissivity is $\frac{x}{8}$ where $x=$ __________
(Given $\sigma=6.0 \times 10^{-8} \mathrm{~W} \mathrm{~m}^{-2} \mathrm{~K}^{-4}, \pi=3.14$ and assume that the emissivity of wire material is same at all wavelength.)
Answer
5
$\mathrm{L}=10 \mathrm{~cm}, \mathrm{~d}=0.5 \mathrm{~mm}, \mathrm{~T}=1727^{\circ} \mathrm{C}=2000 \mathrm{~K}$
Power, $\mathrm{P}=94.2 \mathrm{~W}$
$\mathrm{P}=\varepsilon \sigma \mathrm{AT}^{4}$
$94.2=\varepsilon \times\left(6 \times 10^{-8}\right)(\pi \mathrm{dL})(2000)^{4}$
$94.2=\varepsilon \times\left(6 \times 10^{-8}\right)(3.14)(0.5)\left(10^{-3}\right)$
$\quad$$\quad$$\quad$$\quad$$\left(10 \times 10^{-2}\right)(2000)^{4}$
$\varepsilon=\frac{94.2}{(94.2)(16)}=\frac{5}{8}$
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Question 34 Marks
A container contains a liquid with refractive index of 1.2 up to a height of 60 cm and another liquid having refractive index 1.6 is added to height H above first liquid. If viewed from above, the apparent shift in the position of bottom of container is 40 cm. The value of H is __________ cm.
(Consider liquids are immisible)
Answer
80
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$y=$ apparent depth of bottom
$\frac{\mathrm{y}}{1}=\frac{\mathrm{H}}{1.6}+\frac{60}{1.2}$
Shift $=40$
$H+60-y=40$
$H+60-\frac{H}{1.6}-\frac{60}{1.2}=40$
$\frac{6}{16} H=30$
$\mathrm{H}=80 \mathrm{~cm}$
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Question 44 Marks
For ac circuit shown in figure, $\mathrm{R}=100$ $\mathrm{k} \Omega$ and $\mathrm{C}=100$ $\mathrm{pF}$ and the phase difference between $\mathrm{V}_{\text {in}}$ and $\left(V_{B}-V_{A}\right)$ is $90^{\circ}$. The input signal frequency is $10^{x}$ $\mathrm{rad} / \mathrm{sec}$, where 'x' is
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Answer
5
Input voltage
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$V_{A}-V_{B}:$
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$\theta+\theta=90^{\circ} ; \theta=45^{\circ}$
$\tan \theta=\frac{\mathrm{X}_{\mathrm{C}}}{\mathrm{R}}$
$\mathrm{X}_{\mathrm{C}}=\mathrm{R} \Rightarrow \frac{1}{\mathrm{~W}_{\mathrm{C}}}=\mathrm{R}$
$\mathrm{W}=\frac{1}{\mathrm{R}_{\mathrm{C}}}=\frac{1}{100 \times 10^{3} \times 100 \times 10^{-12}}$
$=\frac{10^{12}}{10^{7}}=10^{5}$
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Question 54 Marks
$A, B$ and $C$ are disc, solid sphere and spherical shell respectively with same radii and masses. These masses are placed as shown in figure.
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The moment of inertia of the given system about $P Q$ is $\frac{x}{15} I$, where $I$ is the moment of inertia of the disc about its diameter. The value of $x$ is __________ .
Answer
199
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All bodies have same mass and same radius.
$\mathrm{A} \rightarrow$ Disc
B $\rightarrow$ Solid sphere
$\mathrm{C} \rightarrow$ Spherical shell
and, $\mathrm{I}=\frac{\mathrm{MR}^{2}}{4}$
$\mathrm{I}_{\mathrm{PQ}}=\frac{\mathrm{MR}^{2}}{4}+\left(\frac{2}{5} \mathrm{MR}^{2}+\mathrm{MR}^{2}\right)+\left(\frac{2}{3} \mathrm{MR}^{2}+\mathrm{MR}^{2}\right)$
$\mathrm{I}_{\mathrm{PQ}}=\frac{15 \mathrm{MR}^{2}+24 \mathrm{MR}^{2}+60 \mathrm{MR}^{2}+40 \mathrm{MR}^{2}+60 \mathrm{MR}^{2}}{60}$
$\mathrm{I}_{\mathrm{PQ}}=\frac{199}{60} \mathrm{MR}^{2}=\frac{199}{15}\left(\frac{\mathrm{MR}^{2}}{4}\right)$
$=\frac{199}{15} \mathrm{I}$
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