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#### As shown in the figure, in Young's double slit experiment, a thin plate of thickness$t=10\mu m$ and refractive index $t=1.2$ is inserted infront of slit $S_{1}$. The experiment is conducted in air$(\mu =1)$ and uses a monochromatic light of wavelength $\lambda =500\; nm$. Due to the insertion of the plate, centralmaxima is shifted by a distance of $x\beta _{0}.\beta _{0}$ is the fringe-width befor the insertion of the plate. The value of the x is ________.Option: 1 4Option: 2 __Option: 3 __Option: 4 __

$Given\; t=10 \times 10^{-6} \mathrm{~m} \mu=1.2$

$\lambda=500 \times 10^{-9} \mathrm{~m}$

When the glass slab inserted infront of one slit then the shift of central fringe is obtained by

\begin{aligned} & \mathrm{t}=\frac{\mathrm{n} \lambda}{(\mu-1)} \\ & \Rightarrow \quad 10 \times 10^{-6}=\frac{\mathrm{n} \times 500 \times 10^{-9}}{(1.2-1)} \\ & 10 \times 10^{-6}=\frac{\mathrm{n} \times 500 \times 10^{-9}}{0.2} \\ & \mathrm{n}=4 \end{aligned}

#### A square shaped coil of area$70\; cm^{2}$ having 600 turns rotates in a magnetic field of $0.4 \mathrm{wbm}^{-2}$, about an axis which is parallel to one of the side of the coil and perpendicular to the direction of field. If the coil completes 500 revolution in a minute, the instantaneous emf when the plane of the coil is inclined at $60^{\circ}$ with the field, will be ________ V.$\left(\text { Take } \pi=\frac{22}{7}\right)$Option: 1 44Option: 2 __Option: 3 __Option: 4 __

$\text { Area (A) }=70 \mathrm{~cm}^2=70 \times 10^{-4} \mathrm{~m}^2$

\begin{aligned} & \mathrm{B}=0.4 \mathrm{~T} \\ & \mathrm{f}=\frac{500 \text { revolution }}{60 \text { minute }}=\frac{500}{60} \frac{\mathrm{rev} .}{\mathrm{sec} .} \end{aligned}

Induced emf in rotating coil is given by

\begin{aligned} & \mathrm{e}=\mathrm{N} \omega \mathrm{BA} \sin \theta \\ & =600 \times 2 \times \frac{22}{7} \times \frac{500}{60} \times 0.4 \times 70 \times 10^{-4} \sin 30^{\circ} \\ & =600 \times 2 \times \frac{22}{7} \times \frac{500}{6} \times 0.4 \times 70 \times 10^{-4} \times \frac{1}{2} \\ & =44 \text { Volt } \end{aligned}

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#### A block is fastened to a horizontal spring. The block is pulled to a distance x = 10 cm from its equilibrium position (at x = 0 ) on a frictionless surface from rest. The energy of the block at $x = 5 cm$ $0.25 J$. The spring constant of the spring is ________$Nm^{-1}$Option: 1 50Option: 2 ___Option: 3 __Option: 4 __

Given

$A=10cm$

At any instant total energy for free oscillation remains constant$=\frac{1}{2} \mathrm{kA}^2$

\begin{aligned} & \Rightarrow \frac{1}{2} \mathrm{kA}^2=0.25 \mathrm{~J} \\ & \Rightarrow \frac{1}{2} \mathrm{kA}^2=0.25 \mathrm{~J} \quad \Rightarrow \mathrm{K}=\frac{0.25 \times 2}{\mathrm{~A}^2} \\ & \Rightarrow \mathrm{k}=\frac{0.50}{(10 \mathrm{~cm})^2}=\frac{0.50}{\left(10 \times 10^{-2}\right)}=\frac{0.50 \times 10^4}{100} \\ & \mathrm{k}=0.50 \times 100=50 \mathrm{~N} / \mathrm{m} \end{aligned}

#### For a body projected at an angle with the horizontal from the ground, choose the correct statement.Option: 1 The vertical component of momentum is maximum at the highest point.Option: 2 The Kinetic Energy (K.E.) is zero at the highest point of projectile motion.Option: 3 The horizontal component of velocity is zero at the highest point.Option: 4 Gravitational potential energy is maximum at the highest point.

At highest point height is maximum and vertical component of velocity is zero.
So momentum is zero. At highest point horizontal component of velocity will not be zero but vertical component of velocity is equal to zero and because of this K.E. will not be equal to zero. Gravitational potential energy is maximum at highest point and equal to$\mathrm{mgH}=\mathrm{mg}\left(\frac{\mathrm{u}^2 \sin ^2 \theta}{2 \mathrm{~g}}\right)$

Therefore the correct option is (4).

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#### Nucleus A having $\mathrm{Z=17}$ and equal number of protons and neutrons has $\mathrm{1.2 \mathrm{MeV}}$ binding energy per nucleon. Another nucleus $\mathrm{\mathrm{B}\: of\: Z=12}$ has total 26 nucleons and $1.8 \mathrm{MeV}$ binding energy per nucleons. The difference of binding energy of $\mathrm{B \: and \: A}$ will be ___________$\mathrm{\mathrm{MeV}.}$  Option: 1 6Option: 2 -Option: 3 -Option: 4 -

$\mathrm{ \text { For Nucleus A } }$
$\mathrm{Z}=17=\text { Nummber of protons }$           $\mathrm{Given\left ( Z=N \right )\therefore N=17}$
$\mathrm{A}=34=\mathrm{Z}+\mathrm{N}$
$\mathrm{E}_{\mathrm{bn}}=1.2 \mathrm{MeV}$
$\frac{\mathrm{\left(\mathrm{E}_{\mathrm{B}}\right)_1 }}{A}=1.2 \mathrm{MeV}$
$\left(\mathrm{E}_{\mathrm{B}}\right)_1=(1.2 \mathrm{MeV}) \times \mathrm{A}$
$\left(\mathrm{E}_{\mathrm{B}}\right)_1=(1.2 \mathrm{MeV}) \times 34$
$\left(\mathrm{E}_{\mathrm{B}}\right)_1=40.8 \mathrm{MeV} \rightarrow \text { Binding energy of Nucleus } \mathrm{A}$

#### In the given circuit, the value of $\mathrm{\left|\frac{\mathrm{I}_1+\mathrm{I}_3}{\mathrm{I}_2}\right|}$ is ___________.Option: 1 2Option: 2 -Option: 3 -Option: 4 -

Apply KVL in loop (1)
$\mathrm{ 20-10-10 \mathrm{I}=0 }$
Or $\mathrm{ I=1 \mathrm{Amp} }$
Apply KVL in loop (2)

$\mathrm{-10+5 I^{\prime}=0 }$
$\mathrm{ \text { Or } I^{\prime}=2 \mathrm{Amp} }$
On comparing $\mathrm{I}_3=1 \mathrm{~A}$
$\mathrm{I}_2=\mathrm{I}_1=\frac{\mathrm{I}^{\prime}}{2}=1 \mathrm{Amp}$
So, the value of $\left|\frac{\mathrm{I}_1+\mathrm{I}_3}{\mathrm{I}_2}\right|=\left|\frac{1+1}{1}\right|=2 \mathrm{Amp}.$

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#### The surface of water in a water tank of cross section area $750 \mathrm{~cm}^2$ on the top of a house is $\mathrm{h \mathrm{~m}}$ above the tap level. The speed of water coming out through the tap of cross section area $500 \mathrm{~mm}^2 is 30 \mathrm{~cm} / \mathrm{s}$. At that instant, $\mathrm{\frac{d h}{d t}}$ is $\mathrm{x\times 10^{-3} \mathrm{~m} / \mathrm{s}}$. The value of $\mathrm{x}$ will be_______________  Option: 1 2Option: 2 -Option: 3 -Option: 4 -

By using equation of continuity
\begin{aligned} & \mathrm{A}_1 \mathrm{v}_1=\mathrm{A}_2 \mathrm{v}_2 \\ & 750 \times 10^{-4} \times \mathrm{v}_1=500 \times 10^{-6} \times 30 \times 10^{-2} \\ & \mathrm{v}_1=20 \times 10^{-4} \mathrm{~m} / \mathrm{sec} \\ & \mathrm{v}_1=2 \times 10^{-3} \mathrm{~m} / \mathrm{sec} \end{aligned}
Given : $\frac{\mathrm{dh}}{\mathrm{dt}}=\mathrm{v}=\mathrm{x} \times 10^{-3} \mathrm{~m} / \mathrm{sec}.$
Therefore

$\mathrm{x}=2$

#### The escape velocities of two planets A and B are in the ratio 1: 2. If the ratio of their radii respectively is 1: 3, then the ratio of acceleration due to gravity of planet A to the acceleration of gravity of planet B will be :Option: 1 $\frac{3}{2}$Option: 2 $\frac{2}{3}$Option: 3 $\frac{3}{4}$Option: 4 $\frac{4}{3}$

Given

\begin{aligned} & \frac{\mathrm{v}_{\mathrm{A}}}{\mathrm{v}_{\mathrm{B}}}=\frac{1}{2} \\ & \frac{\mathrm{r}_{\mathrm{A}}}{\mathrm{r}_{\mathrm{B}}}=\frac{1}{3} \\ & \frac{\mathrm{g}_{\mathrm{A}}}{\mathrm{g}_{\mathrm{B}}}=? \end{aligned}

As we know,

$\mathrm{v}=\sqrt{\frac{2 \mathrm{GM}}{\mathrm{R}}}$

Hence,

$\frac{\mathrm{v}_{\mathrm{A}}}{\mathrm{v}_{\mathrm{B}}}=\frac{\sqrt{\frac{2 \mathrm{GM}_{\mathrm{A}}}{\mathrm{R}_{\mathrm{A}}}}}{\sqrt{\frac{2 \mathrm{GM}_{\mathrm{B}}}{\mathrm{R}_{\mathrm{B}}}}}=\sqrt{\frac{\mathrm{M}_{\mathrm{A}} \mathrm{R}_{\mathrm{B}}}{\mathrm{M}_{\mathrm{B}} \mathrm{R}_{\mathrm{A}}}}=\frac{1}{2}$ _____________(i)

Given

$\frac{\mathrm{R}_{\mathrm{A}}}{\mathrm{R}_{\mathrm{B}}}=\frac{1}{3}:$__________(ii)

Therefore,

\begin{aligned} \frac{g_{\mathrm{A}}}{g_{\mathrm{B}}} & =\frac{\mathrm{M}_{\mathrm{A}} \mathrm{R}_{\mathrm{A}}^2}{\mathrm{M}_{\mathrm{B}} \mathrm{R}_{\mathrm{B}}^2} \\ & =\frac{1}{4} \times \frac{1}{3} \times 9 \\ & =\frac{3}{4} \end{aligned}

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#### A coil is placed in magnetic field such that plane of coil is perpendicular to the direction of magnetic field. The magnetic flux through a coil can be changed:A. By changing the magnitude of the magnetic field within the coil.B. By changing the area of coil within the magnetic field.C. By changing the angle between the direction of magnetic field and the plane of the coil.D. By reversing the magnetic field direction abruptly without changing its magnitude.Choose the most appropriate answer from the options given below :Option: 1 A and B onlyOption: 2 A, B and D onlyOption: 3 A, B and C onlyOption: 4 A and C only

$\phi=\mathrm{BA} \cos \theta$
This show
(1) by changing $B$
(2) by changing $\mathrm{A}$
(3) Angle $(\theta)$ between B and plane of coil.

#### A force $\mathrm{\mathrm{F}=\left(5+3 y^2\right)}$ acts on a particle in the $\mathrm{y}$-direction, where $\mathrm{F}$ is in newton and $\mathrm{y}$ is in meter. The work done by the force during a displacement from $\mathrm{y=2 \mathrm{~m} \: to\: y=5 \mathrm{~m}\: is \: }$_______________$\mathrm{J}$  Option: 1 132 JOption: 2 -Option: 3 -Option: 4 -

Given :
$\mathrm{F=\left(5+3 y^2\right)}$ in the $\mathrm{ y}$ direction
Work done is given by
$\mathrm{W}=\int_{\mathrm{y}_1}^{\mathrm{y}_2} \text { F.dy }$
$\mathrm{y}_1=2 \mathrm{~m}, \quad \mathrm{y}_2=5 \mathrm{~m}$
$\mathrm{~W}=\int_2^5\left(5+3 \mathrm{y}^2\right) \mathrm{dy}$
$\mathrm{W}=\int_2^5 5 \mathrm{dy}+\int_2^5 3 \mathrm{y}^2 \mathrm{dy}$
$\mathrm{W}=[5 \mathrm{y}]_2^5+\left[\frac{3 \mathrm{y}^3}{3}\right]_2^5$
$\mathrm{~W}=(5 \times 5-5 \times 2)+(125-8)$
$\mathrm{W}=(25-10)+117$
$\mathrm{W}=132 \mathrm{Joule}$