datavorous/entrance-exam-dataset huggingface dataset:
A spherically symmetric gravitational system of particles has a mass density $\rho=\left\{\begin{array}{ll}\rho_{0} & \text { for } \mathrm{r} \leq \mathrm{R} \\ 0 & \text { for } \mathrm{r}>\mathrm{R}\end{array}\right.$

where $\mathrm{r}_{0}$ is a constant. A test mass can undergo

circular motion under the influence of the gravitational field of particles. Its speed $\mathrm{V}$ as a function of distance $r(0

PhysicsElectrostaticsBITSATBITSAT 2014

Options:

A
B
C
D

Solution:

1141 Upvotes

Verified Answer

The correct answer is:

Given mass density \(P=\left\{P_{0}\right.\) for \(r \leq R O\) for \(\left.r>R\right\}\)

Since gravitational depends on mass, which is zero for \(r>R\), hence it only acts for \(r \leq R\).

\(\begin{aligned}

&\overrightarrow{\mathrm{F}}_{\text {graw }}=\overrightarrow{\mathrm{F}}_{\text {circular }} \\

&\Rightarrow \frac{\mathrm{GM}_{\text {sphere }} \times \mathrm{M}}{\mathrm{R}^{2}}=\frac{\mathrm{M}(\mathrm{e})^{2}}{\mathrm{R}} \\

&\Rightarrow \mathrm{V}^{2} \alpha \frac{\mathrm{M}_{\text {sphere }}}{\mathrm{R}} \text { since } \mathrm{r} \leq \mathrm{R} \\

&\Rightarrow \mathrm{V}^{2} \alpha \frac{\mathrm{R}^{3}}{\mathrm{R}} \alpha \mathrm{R}^{2} \Rightarrow \mathrm{V} \alpha \mathrm{R} \quad \text { straight line }+\mathrm{ve} \text { slope }

\end{aligned}\)

Beyond \(\mathrm{r}=\mathrm{R}\), mass density \(=0 \Rightarrow\) acceleration due to gravity \(=0 \Rightarrow\) velocity stays constant.

Looking for more such questions to practice?

Download the MARKS App - The ultimate prep app for IIT JEE &
NEET with chapter-wise PYQs, revision notes, formula sheets,
custom tests & much more.

Use on iOS or Desktop Download from Google Play

Options:
A
B
C
D

The correct answer is:

Solution:
Given mass density \(P=\left\{P_{0}\right.\) for \(r \leq R O\) for \(\left.r>R\right\}\)

Since gravitational depends on mass, which is zero for \(r>R\), hence it only acts for \(r \leq R\).

\(\begin{aligned}

&\overrightarrow{\mathrm{F}}_{\text {graw }}=\overrightarrow{\mathrm{F}}_{\text {circular }} \\

&\Rightarrow \frac{\mathrm{GM}_{\text {sphere }} \times \mathrm{M}}{\mathrm{R}^{2}}=\frac{\mathrm{M}(\mathrm{e})^{2}}{\mathrm{R}} \\

&\Rightarrow \mathrm{V}^{2} \alpha \frac{\mathrm{M}_{\text {sphere }}}{\mathrm{R}} \text { since } \mathrm{r} \leq \mathrm{R} \\

&\Rightarrow \mathrm{V}^{2} \alpha \frac{\mathrm{R}^{3}}{\mathrm{R}} \alpha \mathrm{R}^{2} \Rightarrow \mathrm{V} \alpha \mathrm{R} \quad \text { straight line }+\mathrm{ve} \text { slope }

\end{aligned}\)

Beyond \(\mathrm{r}=\mathrm{R}\), mass density \(=0 \Rightarrow\) acceleration due to gravity \(=0 \Rightarrow\) velocity stays constant.

Topic: Physics, Electrostatics, BITSAT, BITSAT 2014
A spherically symmetric gravitational systom of particles has mass density ρ=ρ0 for r≤R0 for r>R where, ρ0 is a constant. A test mass can undergo circular motion under the influence of the gravitational field of particles. Its speed v as a function of distance r(0<r<∞) from the centre of the system is represented by

Options:
A
B
C
D

The correct answer is:

Solution:
When r≤R, then force on the test mass n at the surface of the sphere =mg

Force on the test mass at distance r from the centre of sphere is

If r<R, then F=GMmR2·rR=GMmR3·r

∴mv2r=GMmR3·r

∴v α r

If r>R then F=GMmr2

∴Mv2r=GMmr2

∴ v α 1r

Topic: Physics, Gravitation, JEE Main
A spherically symmetric gravitational systom of particles has mass density ρ=ρ0 for r≤R0 for r>R where, ρ0 is a constant. A test mass can undergo circular motion under the influence of the gravitational field of particles. Its speed v as a function of distance r(0<r<∞) from the centre of the system is represented by

Options:
A
B
C
D

The correct answer is:

Solution:
When r≤R, then force on the test mass n at the surface of the sphere =mg

Force on the test mass at distance r from the centre of sphere is

If r<R, then F=GMmR2·rR=GMmR3·r

∴mv2r=GMmR3·r

∴v α r

If r>R then F=GMmr2

∴Mv2r=GMmr2

∴ v α 1r

Topic: Physics, Gravitation, BITSAT, BITSAT 2018
A spontaneous reaction is impossible if

Options:
A both $\Delta \mathrm{H}$ and $\Delta \mathrm{S}$ are negative
B both $\Delta \mathrm{H}$ and $\Delta \mathrm{S}$ are positive
C $\Delta \mathrm{H}$ is negative and $\Delta \mathrm{S}$ is positive
D $\Delta \mathrm{H}$ is positive and $\Delta \mathrm{S}$ is negative

The correct answer is:
$\Delta \mathrm{H}$ is positive and $\Delta \mathrm{S}$ is negative

Solution:
$\Delta \mathrm{G}=\Delta \mathrm{H}-\mathrm{T} \Delta \mathrm{S} ; \Delta \mathrm{G}$ is positive for a reaction

to be non-spontaneous when $\Delta \mathrm{H}$ is positive and $\Delta \mathrm{S}$ is negative.

Topic: Chemistry, Thermodynamics (C), BITSAT, BITSAT 2015
A spring is compressed between two toy carts of mass $\mathrm{m}_{1}$ and $\mathrm{m}_{2}$. When the toy carts are released, the springs exert equal and opposite average forces for the same time on each toy cart. If $v_{1}$ and $\mathrm{v}_{2}$ are the velocities of the toy carts and there is no friction between the toy carts and the ground, then :

Options:
A $\mathrm{v}_{1} / \mathrm{v}_{2}=\mathrm{m}_{1} / \mathrm{m}_{2}$
B $\mathrm{v}_{1} / \mathrm{v}_{2}=\mathrm{m}_{2} / \mathrm{m}_{1}$
C $\mathrm{v}_{1} / \mathrm{v}_{2}=-\mathrm{m}_{2} / \mathrm{m}_{1}$
D $\mathrm{v}_{1} / \mathrm{v}_{2}=-\mathrm{m}_{1} / \mathrm{m}_{2}$

The correct answer is:
$\mathrm{v}_{1} / \mathrm{v}_{2}=-\mathrm{m}_{2} / \mathrm{m}_{1}$

Solution:
Applying law of conservation of linear momentum
$\mathrm{m}_{1} \mathrm{v}_{1}+\mathrm{m}_{2} \mathrm{v}_{2}=0$
$\frac{\mathrm{m}_{1}}{\mathrm{~m}_{2}}=-\frac{\mathrm{v}_{2}}{\mathrm{v}_{1}}$ or $\frac{\mathrm{v}_{1}}{\mathrm{v}_{2}}=-\frac{\mathrm{m}_{2}}{\mathrm{~m}_{1}}$

Topic: Physics, Center of Mass Momentum and Collision, BITSAT, BITSAT 2010
A square gate of size $1 \mathrm{~m} \times 1 \mathrm{~m}$ is hinged at its mid-point. A fluid of density $\rho$ fills the space to the left of the gate. The force F required to hold the gate stationary is

Options:
A $\frac{\rho g}{3}$
B $\frac{\rho g}{2}$
C $\frac{\rho g}{6}$
D $\frac{\rho g}{8}$

The correct answer is:
$\frac{\rho g}{6}$

Solution:
The net force acting on the gate element of width dy at a depth y from the surface of the fluid, is

$$

\begin{aligned}

\mathrm{dy} &=\left(\mathrm{p}_{0}+\rho_{\mathrm{g}} \mathrm{y}-\mathrm{p}_{0}\right) \times 1 \mathrm{dy} \\

&=\rho \mathrm{gydy}

\end{aligned}

$$

Torque about the hinge is

$$

\mathrm{d} \tau=\operatorname{pgydy} \times\left(\frac{1}{2}-\mathrm{y}\right)

$$

Net torque experience. by the gate is

$$

\begin{aligned}

\tau_{\text {net }} &=\int \mathrm{d} \tau+\mathrm{F} \times \frac{1}{2} \\

&=\int_{0}^{1} \rho g \mathrm{ydy}\left(\frac{1}{2}-\mathrm{y}\right)+\mathrm{F} \times \frac{1}{2}=0 \\

\Rightarrow & \mathrm{F}=\frac{\rho \mathrm{g}}{6}

\end{aligned}

$$

i.e., The force $F$ required to hold the gate

stationary is $\frac{p g}{a}$

Topic: Physics, Rotational Motion, BITSAT, BITSAT 2015
A steady current is set up in a cubic network composed of wires of equal resistance and length d as shown in figure. What is the magnetic field at the centre $\mathrm{P}$ due to the cubic network

Options:
A $\frac{\mu_{0}}{4 \pi} \frac{2 I}{d}$
B $\frac{\mu_{0}}{4 \pi} \frac{2 \mathrm{I}}{\sqrt{2} \mathrm{~d}}$
C 0
D $\frac{\mu_{0}}{4 \pi} \frac{\theta \pi \mathrm{I}}{\mathrm{d}}$

The correct answer is:
0

Solution:
Bysymmetry, the magnetic field at the centre Pis zero.

Topic: Physics, Magnetic Effects of Current, BITSAT, BITSAT 2011
A steel wire of cross-sectional area 4 cm2 has elastic limit of 2.2×108 N m-2. The maximum upward acceleration that can be given to a 1000 kg elevator supported by this steel wire if the stress is to exceed one-fourth of the elastic limit is [Take, g=10 m s-2]

Options:
A 10 m s-2
B 9 m s-2
C 15 m s-2
D 12 m s-2

The correct answer is:
12 m s-2

Solution:
Given, Y=2.2×108 m s-2

Area of cross-section of steel wire,

A=4 cm2=4×10-4 m2

Mass of elevator, m=1000 kg

The maximum tension in the steel wire that can support elevator is given by

T=14× stress × area of cross-section

=14×2.2×108×4×10-4

=2.2×104 N m-2

If f be the maximum upward acceleration of the elevator, then

T=m g+f

2.2×104=1000 10+f

⇒ f=12 m s-2

Topic: Physics, Mechanical Properties of Solids, BITSAT, BITSAT 2019
A stone is projected with velocity 2gh, so that it just clears two walls of equal height h, at distance of 2h from each other. The time interval of passing between the two walls is

Options:
A hg
B 2hg
C 2hg
D 2hg

The correct answer is:
2hg

Solution:
2h=ut

ux=2hΔt ∵u=ux,t=Δt....i

h=uyt-12gt2

gt2-2uyt+2h=0

t1=2uy+4uy2-8gh2g

t2=2uy-4uy2-8gh2g

Δt=t1-t2=4uy2-8ghg

uy2=g2(Δt)24+2gh

ux2+uy2=u2=(2gh)2

4n2(Δt)2+g2(Δt)24+2gh=4gh

g24Δt4-2ghΔt2+4h2=0

Δt=2gh±4g2h2-4g2h2g2/2=4hg

Δt=2hg

Topic: Physics, Motion In One Dimension, BITSAT, BITSAT 2018
A stone projected with a velocity a at an angle $\theta$ with the horizontal reaches maximum height $\mathrm{H}_{1}$. When it is projected with velocity u at an angle $\left(\frac{\pi}{2}-\theta\right)$ with the horizontal, it reaches maximum height $\mathrm{H}_{2}$. The relation between the horizontal range $\mathrm{R}$ of the projectile, heights $\mathrm{H}_{1}$ and $\mathrm{H}_{2}$ is

Options:
A $\mathrm{R}=4 \sqrt{\mathrm{H}_{1} \mathrm{H}_{2}}$
B $\mathrm{R}=4\left(\mathrm{H}_{1}-\mathrm{H}_{2}\right)$
C $\mathrm{R}=4\left(\mathrm{H}_{1}+\mathrm{H}_{2}\right)$
D $\mathrm{R}=\frac{\mathrm{H}_{1}^{2}}{\mathrm{H}_{2}^{2}}$

The correct answer is:
$\mathrm{R}=4 \sqrt{\mathrm{H}_{1} \mathrm{H}_{2}}$

Solution:
$\begin{aligned} & \mathrm{H}_{1}=\frac{\mathrm{u}^{2} \sin ^{2} \theta}{2 \mathrm{~g}} \\ \text { and } \mathrm{H}_{2}=\frac{\mathrm{u}^{2} \sin ^{2}\left(90^{\circ}-\theta\right)}{2 \mathrm{~g}}=\frac{\mathrm{u}^{2} \cos ^{2} \theta}{2 \mathrm{~g}} \\ \mathrm{H}_{1} \mathrm{H}_{2}=& \frac{u^{2} \sin ^{2} \theta}{2 \mathrm{~g}} \times \frac{u^{2} \cos ^{2} \theta}{2 \mathrm{~g}}=\frac{\left(u^{2} \sin 2 \theta\right)^{2}}{16 g^{2}}=\frac{R^{2}}{16} \\ \therefore R=4 \sqrt{H_{1} H_{2}} \end{aligned}$

Topic: Physics, Motion In Two Dimensions, BITSAT, BITSAT 2016
A straight rod of length 9 units slides with its endsA, $B$ always on the $X$ and Y-axis respectively. Then the locus of the centroid of $\Delta \mathrm{OAB}$ is:

Options:
A $x^{2}+y^{2}=3$
B $x^{2}+y^{2}=9$
C $x^{2}+y^{2}=1$
D $x^{2}+y^{2}=81$

The correct answer is:
$x^{2}+y^{2}=9$

Solution:
Let end $A$ is $(a, 0)$ and end $\mathrm{B}$ is $(0,$ b) then

$\mathrm{a}^{2}+\mathrm{b}^{2}=81$

If centroid is $(x, y)$ then

$\mathrm{x}=\frac{\mathrm{a}}{3}, \mathrm{y}=\frac{\mathrm{b}}{3},$ putting in (1) we get the locus as

$x^{2}+y^{2}=9$

Topic: Mathematics, Straight Lines, BITSAT, BITSAT 2020
A string of length $\ell$ is fixed at both ends. It is vibrating in its $3^{\text {rd }}$ overtone with maximum amplitude'a'. The amplitude at a distance $\ell / 3$ from one end is

Options:
A a
B 0
C $\frac{\sqrt{3} \mathrm{a}}{2}$
D $\frac{a}{2}$

The correct answer is:
$\frac{\sqrt{3} \mathrm{a}}{2}$

Solution:
For a string vibrating in its $\mathrm{n}^{\text {th }}$ overtone $(\mathrm{n}+1)^{\text {th }}$ harmonic) $\mathrm{y}=2 \mathrm{~A} \sin \left(\frac{(\mathrm{n}+1) \pi \mathrm{x}}{\mathrm{L}}\right) \cos \omega \mathrm{t}$

$$

\begin{array}{l}

\text { For } \left.x=\frac{\ell}{3}\right] 2 A=a \& n=3 \\

y=\left[a \sin \left(\frac{4 \pi}{\ell} \cdot \frac{\ell}{3}\right)\right] \cos \omega t \\

=a \sin \frac{4 \pi}{3} \cos \omega t=-a\left(\frac{\sqrt{3}}{2}\right) \cos \omega t

\end{array}

$$

i.e. at $x=\frac{\ell}{3},$ the amplitude is $\frac{\sqrt{3} a}{2}$

Topic: Physics, Waves and Sound, BITSAT, BITSAT 2016
A student read common difference of an A. P. as

-2 instead of 2 and got the sum of first 5 terms as -5 . Actual sum of first five terms is

Options:
A 25
B -25
C -35
D 35

The correct answer is:
35

Solution:
$\frac{5}{2}\{2 \mathrm{a}+4(-2)\}=-5 \Rightarrow 2 \mathrm{a}=6$

$\therefore$ Actual sum $=\frac{5}{2}\{6+4(2)\}=35$

Topic: Mathematics, Sequences and Series, BITSAT, BITSAT 2020
A submarine $A$ travelling at $17 \mathrm{~m} / \mathrm{s}$ is being chased along the line of its velocity by another submarine $B$ travelling at $34 \mathrm{~m} / \mathrm{s}$. B sends a sonar signal of $600 \mathrm{~Hz}$ to detect $A$ and receives a reflected sound of frequency $v$. The of $v$ is
[Speed of sound in water $=1500 \mathrm{~ms}^{-1}$ ]

Options:
A
$613.7 \mathrm{~Hz}$

B
$6137 \mathrm{~Hz}$

C
$62 \mathrm{~Hz}$

D
$539 \mathrm{~Hz}$

The correct answer is:

$613.7 \mathrm{~Hz}$

Solution:
$\therefore$ Given, velocity of submarine $A, v_A=17 \mathrm{~m} / \mathrm{s}$
Velocity of submarine $B, v_B=34 \mathrm{~m} / \mathrm{s}$
Signal sent by submarine $B$ is detected by submarine $A$ can be shown as

Freqency of the signal, $f_0=600 \mathrm{~Hz}$
Speed of sound in water $v_s=1500 \mathrm{~ms}^{-1}$
Frequency received by submarine $A$ is
$$
f_1=\left(\frac{v_{\mathrm{S}}-v_A}{v_{\mathrm{S}}-v_B}\right) f_0=\left[\frac{1500-17}{1500-34}\right] \times 600 \mathrm{~Hz}
$$
$$
f_1=\frac{1483}{1466} \times 600
$$
Frequency received by submarine $B$ is
$$
f_2=\left(\frac{v_{\mathrm{S}}+v_B}{v_s+v_A}\right) f_1
$$
Substituting given values and $f_1$ value from Eq. (i), we get
$$
\begin{aligned}
f_2 & =\left(\frac{1500+34}{1500+17}\right) \times\left[\frac{1483}{1466} \times 600\right] \\
f_2 & =1.0112 \times 1.0115 \times 600 \\
f_2 & =613.7 \mathrm{~Hz}
\end{aligned}
$$

Topic: Physics, Waves and Sound, BITSAT, BITSAT 2022
A system consists of three particles, each of mass $\mathrm{m}$ and located at (1,1),(2,2) and (3,3) . The co-ordinates of the centre of mass are

Options:
A (1,1)
B (2,2)
C (3,3)
D (6,6)

The correct answer is:
(2,2)

Solution:
The coordinates of C.M of three particle are
\begin{array}{c}
\mathrm{x}=\frac{\mathrm{m}_{1} \mathrm{x}_{1}+\mathrm{m}_{2} \mathrm{x}_{2}+\mathrm{m}_{3} \mathrm{x}_{3}}{\mathrm{~m}_{1}+\mathrm{m}_{2}+\mathrm{m}_{3}} \& \mathrm{y}=\frac{\mathrm{m}_{1} \mathrm{y}_{1}+\mathrm{m}_{2} \mathrm{y}_{2}+\mathrm{m}_{3} \mathrm{y}_{3}}{\mathrm{~m}_{1}+\mathrm{m}_{2}+\mathrm{m}_{3}} \\
\text { here } \mathrm{m}_{1}=\mathrm{m}_{2}=\mathrm{m}_{3}=\mathrm{m} \\
\text { so } \mathrm{x}=\frac{\left(\mathrm{x}_{1}+\mathrm{x}_{2}+\mathrm{x}_{3}\right) \mathrm{m}}{\mathrm{m}+\mathrm{m}+\mathrm{m}}=2
\end{array}
$$
y=\frac{\left(y_{1}+y_{2}+y_{3}\right) m}{m+m+m}=2
$$
so coordinates of $\mathrm{C} . \mathrm{M}$. of three particle are (2,2)
$\mathrm{V}=300 \mathrm{~m} / \mathrm{s}$

Topic: Physics, Center of Mass Momentum and Collision, BITSAT, BITSAT 2010
A tangent drawn to hyperbola x2a2-y2b2=1 at Pπ6 forms a triangle of area 3a2 square units, with coordinate axes. If the eccentricity of hyperbola is e, then the value of e2-9 is

Options:
A 9
B 10
C 11
D 8

The correct answer is:
8

Solution:
The point Pπ6 is asecπ6,btanπ6 or P2a3,b3

∴ Equation of tangent at P is x3a2-y3b=11$

∴ Area of the triangle =12×3a2×3b=3a2

∴ ba=4

e2=1+b2a2=17

Now, e2-9=17-9=8

Hence, option (d) is correct.

Topic: Mathematics, Hyperbola, BITSAT, BITSAT 2018
A thermodynamic system is taken through cyclic process. The total work done in the process is :

Options:
A $100 \mathrm{~J}$
B $300 \mathrm{~J}$
C Zero
D $200 \mathrm{~J}$

The correct answer is:
$300 \mathrm{~J}$

Solution:
Work done $=$ Area under the curve
$\Rightarrow \mathrm{W}=\frac{1}{2} \times(4-2) \times(400-100)=\frac{1}{2}(2) \times 300$
$\mathrm{W}=300 \mathrm{~J}$

Topic: Physics, Thermodynamics, BITSAT, BITSAT 2023 (Memory Based Paper 1)
A thermodynamical system is changed from state $\left(\mathrm{P}_{1}, \mathrm{~V}_{1}\right)$ to $\left(\mathrm{P}_{2}, \mathrm{~V}_{2}\right)$ by two different process, the quantity which will remain same will be

Options:
A $\Delta \mathrm{Q}$
B $\Delta \mathrm{W}$
C $\Delta \mathrm{Q}+\Delta \mathrm{W}$
D $\Delta \mathrm{Q}-\Delta \mathrm{W}$

The correct answer is:
$\Delta \mathrm{Q}-\Delta \mathrm{W}$

Solution:
For all process $\Delta \mathrm{U}=\Delta \mathrm{Q}-\Delta \mathrm{W}$

does not change as it depends on initial final states.

Topic: Physics, Thermodynamics, BITSAT, BITSAT 2013
A thin convergent glass lens $\left(\mu_{g}=1.5\right)$ has a power of $+5.0 \mathrm{D}$. When this lens is immersed in a liquid of refractive index $\mu,$ it acts as a divergent lens of focal length $100 \mathrm{~cm}$. The value of $\mu$ must be

Options:
A $4 / 3$
B $5 / 3$
C $5 / 4$
D $6 / 5$

The correct answer is:
$5 / 3$

Solution:
$\frac{\mathrm{P}_{\mathrm{a}}}{\mathrm{P}_{1}}=\frac{\left(\frac{\mu_{\mathrm{g}}}{\mu_{\mathrm{a}}}-1\right)}{\left(\frac{\mu_{\mathrm{g}}}{\mu_{1}}-1\right)}=\frac{+5}{-100 / 100}=-5$
$-5\left(\frac{\mu_{\mathrm{g}}}{\mu_{1}}-1\right)=\frac{\mu_{\mathrm{g}}}{\mu_{\mathrm{a}}}-1$
$\frac{1.5}{\mu_{1}}-1=\frac{-1}{5}(1.5-1)=-0.1$
$$
\begin{array}{ll}\mu_{1}=\frac{1.5}{0.9}=\frac{5}{3}\end{array}
$$

Topic: Physics, Ray Optics, BITSAT, BITSAT 2011
A thin prism $P_1$ with an angle $6^{\circ}$ and made of glass of refractive index 1.54 is combined with another prism $\mathrm{P}_2$ made from glass of refractive index 1.72 to produce dispersion without average deviation. The angle of prism $\mathrm{P}_2$ is :

Options:
A $6^{\circ}$
B $1.3^{\circ}$
C $7.8^{\circ}$
D $4.5^{\circ}$

The correct answer is:
$4.5^{\circ}$

Solution:
Angle of deviation for first prism
$\delta_1=A_1\left(\mu_1-1\right)=6(1.54-1)$
Angle of deviation for second prism
$\begin{aligned}
& \delta_2=A_2\left(\mu_2-1\right) \\
& =A_2(1.72-1)
\end{aligned}$
For dispersion without deviation
$\begin{aligned}
& \delta_1=\delta_2 \\
& \Rightarrow 6^{\circ}(1.54-1)=\mathrm{A}_2(1.72-1) \\
& \Rightarrow \mathrm{A}_2=\frac{6^{\circ} \times 0.54}{0.72}=\frac{18^{\circ}}{4}=4.5^{\circ}
\end{aligned}$

Topic: Physics, Ray Optics, BITSAT, BITSAT 2023 (Memory Based Paper 2)
A thin rod of length 41 and mass $4 \mathrm{~m}$ is bent at the points as shown in figure. What is the moment of inertia of the rod about the axis passes through point $\mathrm{O}$ and perpendicular to the plane of paper?

Options:
A $\frac{\mathrm{M} l^{2}}{3}$
B $\frac{10 \mathrm{M} l^{2}}{3}$
C $\frac{\mathrm{M} l^{2}}{12}$
D $\frac{\mathrm{M} l^{2}}{24}$

The correct answer is:
$\frac{10 \mathrm{M} l^{2}}{3}$

Solution:
Total moment of inertia

$$

=\mathrm{I}_{1}+\mathrm{I}_{2}+\mathrm{I}_{3}+\mathrm{I}_{4}=2 \mathrm{I}_{1}+2 \mathrm{I}_{2}

$$

$$

\begin{array}{l}

=2\left(l_{1}+l_{2}\right)\left\lceil I_{3}=I_{1}, I_{1}=I_{4}\right] \\

\text { Now, } I_{2}=I_{3}=\frac{M I^{2}}{3}

\end{array}

$$

Using parallel axes theorem, we have

$$

\begin{array}{l}

\mathrm{I}=\mathrm{I}_{\mathrm{CM}}+\mathrm{Mx}^{2} \text { and } \mathrm{x}=\sqrt{l^{2}+\frac{l^{2}}{4}} \\

\mathrm{I}_{1}=\mathrm{I}_{4}=\frac{\mathrm{M} l^{2}}{12}+\mathrm{M}\left[\sqrt{l^{2}+\left(\frac{l}{2}\right)^{2}}\right]^{2}

\end{array}

$$

Putting all values we get Moment of inertia, $\mathrm{I}=10\left(\frac{\mathrm{Ml}^{2}}{3}\right)$

Topic: Physics, Rotational Motion, JEE Main
A thin rod of length 41 and mass $4 \mathrm{~m}$ is bent at the points as shown in figure. What is the moment of inertia of the rod about the axis passes through point $\mathrm{O}$ and perpendicular to the plane of paper?

Options:
A $\frac{\mathrm{M} l^{2}}{3}$
B $\frac{10 \mathrm{M} l^{2}}{3}$
C $\frac{\mathrm{M} l^{2}}{12}$
D $\frac{\mathrm{M} l^{2}}{24}$

The correct answer is:
$\frac{10 \mathrm{M} l^{2}}{3}$

Solution:
Total moment of inertia

$$

=\mathrm{I}_{1}+\mathrm{I}_{2}+\mathrm{I}_{3}+\mathrm{I}_{4}=2 \mathrm{I}_{1}+2 \mathrm{I}_{2}

$$

$$

\begin{array}{l}

=2\left(l_{1}+l_{2}\right)\left\lceil I_{3}=I_{1}, I_{1}=I_{4}\right] \\

\text { Now, } I_{2}=I_{3}=\frac{M I^{2}}{3}

\end{array}

$$

Using parallel axes theorem, we have

$$

\begin{array}{l}

\mathrm{I}=\mathrm{I}_{\mathrm{CM}}+\mathrm{Mx}^{2} \text { and } \mathrm{x}=\sqrt{l^{2}+\frac{l^{2}}{4}} \\

\mathrm{I}_{1}=\mathrm{I}_{4}=\frac{\mathrm{M} l^{2}}{12}+\mathrm{M}\left[\sqrt{l^{2}+\left(\frac{l}{2}\right)^{2}}\right]^{2}

\end{array}

$$

Putting all values we get Moment of inertia, $\mathrm{I}=10\left(\frac{\mathrm{Ml}^{2}}{3}\right)$

Topic: Physics, Rotational Motion, BITSAT, BITSAT 2015
A thin sheet of glass $(\mu=1.5)$ of thickness 6 micron introduced in the path of one of the interfering beams in a double slit experiment shifts the central fringe to a position previously occupied by fifth bright fringe. Then the wavelength of light used is

Options:
A $6000 Ã$
B $3000 Ã$
C $4500 Ã$
D $7500 Ã$

The correct answer is:
$6000 Ã$

Solution:
$\mathrm{n} \lambda=(\mu-1) \mathrm{t} ;$

$\therefore \lambda=\frac{(\mu-1) \mathrm{t}}{\mathrm{n}}=\frac{(1.5-1) \times 6 \times 10^{-6}}{5}=6000 \mathrm{~A}$

Topic: Physics, Wave Optics, BITSAT, BITSAT 2020
A train accelerating uniformly from rest attains a maximum speed of $40 \mathrm{~ms}^{-1}$ in $20 \mathrm{~s}$. It travels at the speed for $20 \mathrm{~s}$ and is brought to rest with uniform retardation in further $40 \mathrm{~s}$. What is the average velocity during the period ?

Options:
A $80 \mathrm{~m} / \mathrm{s}$
B $25 \mathrm{~m} / \mathrm{s}$
C $40 \mathrm{~m} / \mathrm{s}$
D $30 \mathrm{~m} / \mathrm{s}$

The correct answer is:
$25 \mathrm{~m} / \mathrm{s}$

Solution:
(i) $\mathrm{v}=\mathrm{u}+$ at $_{1}$

$$

\begin{array}{l}

40=0+a \times 20 \\

a=2 m / s^{2} \\

v^{2}-u^{2}=2 a s \\

40^{2}-0=2 \times 2 s_{1} \\

\text { (ii) } s_{1}=400 \mathrm{~m} \\

\text { (iii) } v=u+a t \\

\quad 0=40+a \times 40 \\

a=-1 \mathrm{~m} / \mathrm{s}^{2} \\

0^{2}-40^{2}=2(-1) \mathrm{s}_{3} \\

\mathrm{~s}_{3}=800 \mathrm{~m} \\

\text { Total distance travelled }=\mathrm{s}_{1}+\mathrm{s}_{2}+\mathrm{s}_{3} \\

=400+800+800=2000 \mathrm{~m} \\

\text { Total time taken }=20+20+40=80 \mathrm{~s} \\

\text { Average velocity }=\frac{2000}{80}=25 \mathrm{~m} / \mathrm{s}

\end{array}

$$

Topic: Physics, Motion In One Dimension, BITSAT, BITSAT 2014
A train moves towards a stationary observer with speed 34 m s-1. The train blows whistle and its frequency is registered by the observer as ν1. If the train's speed is reduced to 17 m s-1, then the frequency registered is ν2. If the speed of sound is 340 m s-1, then the ratio ν1ν2 is

Options:
A 1918
B 1819
C 2
D 1/2

The correct answer is:
1918

Solution:
According to Doppler's effect, the approximate frequency heard by the stationary observer,

ν=CC-vsν0

Case (I) vs=34 m s-1, where C=Speed of sound in air,

vs=Speed of source, and ν0= frequency of the source.

∴ ν1=340340-34ν0

=340306ν0 ....i

Case (ii) vs=17 m s-1
∴ ν2=340340-17ν0=340323ν0 ....ii

From Eq. (i) and (ii), we get

∴ ν1ν2=340/306340/323=323306=1918

Topic: Physics, Waves and Sound, BITSAT, BITSAT 2018
A transistor has a base current of $1 \mathrm{~mA}$ and emitter current $90 \mathrm{~mA}$. The collector current will

Options:
A $90 \mathrm{~mA}$
B $1 \mathrm{~mA}$
C $89 \mathrm{~mA}$
D $91 \mathrm{~mA}$

The correct answer is:
$89 \mathrm{~mA}$

Solution:
$\quad \mathrm{I}_{\mathrm{C}}=\mathrm{I}_{\mathrm{E}}-\mathrm{I}_{\mathrm{B}}=90-1=89 \mathrm{~mA}$

Topic: Physics, Semiconductors, BITSAT, BITSAT 2010
A transistor is connected in common-emitter (CE) configuration. The collector supply is $8 \mathrm{~V}$ and the voltage drop across a resistor is $500 \Omega$ in the collector circuit is $0.6 \mathrm{~V}$. If the current gain factor $\alpha$ is $0.96$, find the base current

Options:
A
$25 \mu \mathrm{A}$

B
$50 \mu \mathrm{A}$

C
$20 \mu \mathrm{A}$

D
$35 \mu \mathrm{A}$

The correct answer is:

$50 \mu \mathrm{A}$

Solution:
Given, $V_{C C}=8 \mathrm{~V}$
Voltage drop, $V_0=\Delta I_C R_L=0.6 \mathrm{~V}$
Load Resistor, $R_L=500 \Omega$
Gain factor, $\alpha=0.96$
$\because \quad \beta=\frac{\alpha}{1-\alpha}=\frac{0.96}{1-0.96}=24$
Now, $\quad V_0=\Delta I_C R_L$
$0.6=\Delta I_C \times 500$
$\Delta I_C=\frac{0.6}{500}=1.2 \times 10^{-3} \mathrm{~A}$
Now,
$\beta=\frac{\Delta I_C}{\Delta I_B}$
$\therefore$ Base current $\Delta I_B=\frac{\Delta I_C}{\beta}=\frac{1.2 \times 10^{-3}}{24}=5 \times 10^{-5}$
$=50 \times 10^{-6}$
$\Delta I_B=50 \mu \mathrm{A}$

Topic: Physics, Semiconductors, BITSAT, BITSAT 2022