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

Question 14 Marks

Conductor wire ABCDE with each arm 10 cm in length is placed in magnetic field of $\frac{1}{\sqrt{2}}$ Tesla, perpendicular to its plane. When conductor is pulled towards right with constant velocity of $10 \mathrm{~cm} / \mathrm{s}$, induced emf between points A and E is _____________ mV.

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Question 24 Marks

Four capacitor each of capacitance $16 \mu \mathrm{F}$ are connected as shown in the figure. The capacitance between points A and B is : _____________ (in $\mu \mathrm{F}$ ).

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Question 34 Marks

Distance between object and its image (magnified by $-\frac{1}{3}$) is 30 cm. The focal length of the mirror used is $\left(\frac{\mathrm{x}}{4}\right) \mathrm{cm}$,
where magnitude of value of $x$ is _____________.

Answer

$\mathrm{M}=-\frac{1}{3}$
$-\frac{-\mathrm{V}}{-\mathrm{U}}=\frac{-1}{3} \Rightarrow \mathrm{~V}=\frac{\mathrm{U}}{3}$
Distance $\mathrm{b} / \mathrm{w}$ object and image :

$\mathrm{U}-\mathrm{V}=30$
$U-\frac{\mu}{3}=30$
$\Rightarrow \mathrm{U}=45 \quad \mathrm{~V}=15$
$\frac{1}{\mathrm{f}}=\frac{1}{\mathrm{~V}}+\frac{1}{\mathrm{U}}=-\frac{1}{15}-\frac{1}{45}$
$\Rightarrow \mathrm{F}=\frac{45}{4}$
$\mathrm{x}=45$

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Question 44 Marks

Two slabs with square cross section of different materials $(1,2)$ with equal sides $(l)$ and thickness $\mathrm{d}_{1}$ and $\mathrm{d}_{2}$ such that $\mathrm{d}_{2}=2 \mathrm{d}_{1}$ and $l>\mathrm{d}_{2}$. Considering lower edges of these slabs are fixed to the floor, we apply equal shearing force on the narrow faces. The angle of deformation is $\theta_{2}=2 \theta_{1}$. If the shear moduli of material 1 is $4 \times 10^{9} \mathrm{~N} / \mathrm{m}^{2}$, then shear moduli of material 2 is $x \times 10^{9} \mathrm{~N} / \mathrm{m}^{2}$, where value of $x$ is _____________.

Answer

Deformation angle
$2 \theta_{1}=\theta_{2}$
$\Rightarrow 2 \frac{\sigma_{1}}{\eta_{1}}=\frac{\sigma_{2}}{\eta_{2}}$

$\Rightarrow 2\left(\frac{F}{\ell d_{1} \eta_{1}}\right)=\frac{F}{\ell d_{2} \eta_{2}}$
$\Rightarrow \eta_{2}=\frac{\eta_{1}}{4}=1 \times 10^{9} \Rightarrow x=1$

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Question 54 Marks

A circular ring and a solid sphere having same radius roll down on an inclined plane from rest without slipping. The ratio of their velocities when reached at the bottom of the plane is $\sqrt{\frac{x}{5}}$ where $\mathrm{x}=$ _____________.

Answer

Applying Mechanical Energy conservation :
$\mathrm{k}_{\mathrm{i}}+\mathrm{U}_{\mathrm{i}}=\mathrm{k}_{\mathrm{f}}+\mathrm{U}_{\mathrm{f}}$
$\Rightarrow 0+\mathrm{Mgh}=\frac{1}{2} \mathrm{mv}^{2}\left(1+\frac{\mathrm{k}^{2}}{\mathrm{R}^{2}}\right)+0$
$\Rightarrow \mathrm{V}=\sqrt{\frac{2 \mathrm{gh}}{1+\frac{\mathrm{k}^{2}}{\mathrm{R}^{2}}}}$
So Ratio of velocities
$\frac{\mathrm{V}_{\text {Ring }}}{\mathrm{V}_{\text {solids sphere }}}=\sqrt{\frac{1+\frac{2}{5}}{1+1}}=\sqrt{\frac{7}{10}}$
$x=3.5$ Rounding off $x=4$

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JEE SECTION - B [PHYSICS - NUMERIC]

SECTION - B [PHYSICS - NUMERIC]

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