datavorous/entrance-exam-dataset huggingface dataset:
A bag contains $(2 n+1)$ coins. It is known that n of these coins have a head on both sides, whereas the remaining $(\mathrm{n}+1)$ coins are fair. A coin is picked up at random from the bag and tossed. If the probability that the toss results in a head is $31 / 42,$ then $\mathrm{n}$ is equal to

Options:
A 10
B 11
C 12
D 13

The correct answer is:
10

Solution:
Total number of coins $=2 \mathrm{n}+1$

Consider the following events:

$\mathrm{E}_{1}=$ Getting a coin having head on both

sides from the bag.

$\mathrm{E}_{2}=$ Getting a fair coin from the bag

$\mathrm{A}=$ Toss results in a head

Given: $\mathrm{P}(\mathrm{A})=\frac{31}{42}, \mathrm{P}\left(\mathrm{E}_{1}\right)=\frac{\mathrm{n}}{2 \mathrm{n}+1}$

and $\mathrm{P}\left(\mathrm{E}_{2}\right)=\frac{\mathrm{n}+1}{2 \mathrm{n}+1}$

Then,

$\mathrm{P}(\mathrm{A})=\mathrm{P}\left(\mathrm{E}_{1}\right) \mathrm{P}\left(\mathrm{A} / \mathrm{E}_{1}\right)+\mathrm{P}\left(\mathrm{E}_{2}\right) \mathrm{P}\left(\mathrm{A} / \mathrm{E}_{2}\right)$

$\Rightarrow \frac{31}{42}=\frac{n}{2 n+1} \times 1+\frac{n+1}{2 n+1} \times \frac{1}{2}$

$\frac{31}{42}=\frac{n}{2 n+1}+\frac{n+1}{2(2 n+1)}$

$\Rightarrow \frac{31}{42}=\frac{3 n+1}{2(2 n+1)}$

$\Rightarrow \frac{31}{21}=\frac{3 n+1}{2 n+1}$

$n=10$

Topic: Mathematics, Probability, BITSAT, BITSAT 2015
A bag contains 3 red and 3 white balls. Two balls are drawn one by one. The probability that they are of different colours is.

Options:
A $3 / 10$
B $2 / 5$
C $3 / 5$
D None of these

The correct answer is:
$3 / 5$

Solution:
Let $\mathrm{A} \equiv$ event that drawn ball is red $\mathrm{B} \equiv$ event that drawn ball is white Then $\mathrm{AB}$ and $\mathrm{BA}$ are two disjoint cases of the given event.

$$

\begin{array}{l}

\therefore P(A B+B A)=P(A B)+P(B A) \\

=P(A) \quad P\left(\frac{B}{A}\right)+P(B) \quad P\left(\frac{A}{B}\right)

\end{array}

$$

$=\frac{3}{6} \cdot \frac{3}{5}+\frac{3}{6} \cdot \frac{3}{5}=\frac{3}{5}$

Topic: Mathematics, Probability, BITSAT, BITSAT 2015
A bag contains 5 brown and 4 white socks. $A$ man pulls out 2 socks. Find the probability that they are of the same colour.

Options:
A $\frac{4}{9}$
B $\frac{2}{9}$
C $\frac{5}{9}$
D $\frac{7}{9}$

The correct answer is:
$\frac{4}{9}$

Solution:
Let $A \equiv$ event of two socks being brown. $\mathrm{B} \equiv$ event of two socks being white.

$$

\text { Then } \begin{aligned}

& P(A)=\frac{{ }^{5} C_{2}}{{ }^{9} C_{2}}=\frac{5.4}{9.8}=\frac{5}{18}, \mathrm{P}(\mathrm{B}) \\

&=\frac{{ }^{4} \mathrm{C}_{2}}{{ }^{9} \mathrm{C}_{2}}=\frac{4.3}{9.8}=\frac{3}{18}

\end{aligned}

$$

Now, since $\mathrm{A}$ and $\mathrm{B}$ are mutually exclusive events, so required probability

$$

=\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})=\frac{5}{18}+\frac{3}{18}=\frac{4}{9}

$$

Topic: Mathematics, Probability, BITSAT, BITSAT 2014
A bag contains $n+1$ coins. It is known that one of these coins shows heads on both sides, whereas the other coins are fair. One coin is selected at random and tossed. If the probability that toss results in heads is $\frac{7}{12},$ then the value of $n$ is.

Options:
A 3
B 4
C 5
D None of these

The correct answer is:
5

Solution:
Let $E_{1}$ denote the event "a coin with head on both sides is selected" and $\mathrm{E}_{2}$ denotes the event " a fair coin is selected". Let $A$ be

the event " he toss, results in heads".

$$

\begin{array}{l}

\therefore \mathrm{P}\left(\mathrm{E}_{1}\right)=\frac{1}{n+1}, \mathrm{P}\left(\mathrm{E}_{2}\right)=\frac{n}{n+1} \text { and } \\

\mathrm{P}\left(\frac{\mathrm{A}}{\mathrm{E}_{1}}\right)=1, \mathrm{P}\left(\frac{\mathrm{A}}{\mathrm{E}_{2}}\right)=\frac{1}{2} \\

\therefore \mathrm{P}(\mathrm{A})=\mathrm{P}\left(\mathrm{E}_{1}\right) \mathrm{P}\left(\frac{\mathrm{A}}{\mathrm{E}_{1}}\right)+\mathrm{P}\left(\mathrm{E}_{2}\right) \mathrm{P}\left(\frac{\mathrm{A}}{\mathrm{E}_{2}}\right) \\

\Rightarrow \frac{7}{12}=\frac{1}{n+1} \times 1+\frac{n}{n+1} \times \frac{1}{2} \\

\Rightarrow 14 n+14=24+12 n \Rightarrow n=5

\end{array}

$$

Topic: Mathematics, Probability, BITSAT, BITSAT 2014
A ball is dropped vertically from a height d above the ground. It hits the ground and bounces up vertically to a height d/2. Neglecting subsequent motion and air resistance, its velocity v varies with height h above the ground as

Options:
A
B
C
D

The correct answer is:

Solution:
For the uniformly accelerated/decelcrated motion,

v2=u2±2gh

So. from this equation. we can say v-h paragraph is parabola.

Initially, velocity is downwards (-ve) after collision, it reverses the direction with smaller magnitude and velocity is upwards (-ve).

So. graph (a) satisfies these conditions.

Also, When t=0, h=d Velocity increases downwards (U→A)

When t-1 velocity reverses its direction A'→B

Topic: Physics, Motion In One Dimension, BITSAT, BITSAT 2018
A ball of mass $0.2 \mathrm{~kg}$ is thrown vertically upwards by applying a force by hand. If the hand moves $0.2 \mathrm{~m}$ while applying the force and the ball goes upto $2 \mathrm{~m}$ height further, find the magnitude of the force. (Consider $\left.g=10 \mathrm{~m} / \mathrm{s}^{2}\right)$.

Options:
A $4 \mathrm{~N}$
B $16 \mathrm{~N}$
C $20 \mathrm{~N}$
D $22 \mathrm{~N}$

The correct answer is:
$22 \mathrm{~N}$

Solution:
Let the velocity of the ball just when it leaves the hand is $u$ then applying,

$\begin{array}{l}

v^{2}-u^{2}=2 a \text { for upward journey } \\

\Rightarrow-u^{2}=2(-10) \times 2 \\

\Rightarrow u^{2}=40

\end{array}$

Again applying $v^{2}-u^{2}=2 a s$ for the upward journey of the ball, when the ball is in the hands of the thrower,

$\begin{array}{l}

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

\Rightarrow 40-0=2(a) 0.2 \\

\Rightarrow a=100 \mathrm{~m} / \mathrm{s}^{2} \\

\therefore \mathrm{F}=m a=0.2 \times 100=20 \mathrm{~N} \\

\Rightarrow \mathrm{N}-m g=20 \\

\Rightarrow \mathrm{N}=20+2=22 \mathrm{~N}

\end{array}$

Topic: Physics, Laws of Motion, BITSAT, BITSAT 2021
$\quad$ A ball whose density is $0.4 \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3}$ falls into water from a height of $9 \mathrm{~cm}$. To what depth does the ball sink?

Options:
A $2 \mathrm{~cm}$
B $6 \mathrm{~cm}$
C $4.5 \mathrm{~cm}$
D $2.25 \mathrm{~cm}$

The correct answer is:
$6 \mathrm{~cm}$

Solution:
No Solution

Topic: Physics, Mechanical Properties of Fluids, BITSAT, BITSAT 2013
A balloon is filled at 27 °C and 1 atm pressure by 500 m3 He. At -3 °C and 0.5 atm pressure, the volume of He will be

Options:
A 700 m3
B 900 m3
C 1000 m3
D 500 m3

The correct answer is:
900 m3

Solution:
We have, P1V1T1=P2V2T2

or V2=P1V1T2P2T1

=1×500×273-30.5×273+27=1×500×2700.5×300

V2=900 m3

Topic: Physics, Kinetic Theory of Gases, BITSAT, BITSAT 2017
A balloon is rising vertically up with a velocity of $29 \mathrm{~ms}^{-1}$. A stone is dropped from it and it reaches the ground in 10 seconds. The height of the balloon when the stone was dropped from it, was $\left(\mathrm{g}=9.8 \mathrm{~ms}^{-2}\right)$

Options:
A $100 \mathrm{~m}$
B $200 \mathrm{~m}$
C $400 \mathrm{~m}$
D $150 \mathrm{~m}$

The correct answer is:
$200 \mathrm{~m}$

Solution:
For stone to be dropped from rising balloon of velocity $29 \mathrm{~m} / \mathrm{s}, \mathrm{u}=-29 \mathrm{~m} / \mathrm{s}, \mathrm{t}=10 \mathrm{sec} .$

$\begin{array}{l}

\therefore \mathrm{h}=-29 \times 10+\frac{1}{2} \times 9.8 \times 100 \\

=-290+490=200 \mathrm{~m} .

\end{array}$

Topic: Physics, Motion In One Dimension, BITSAT, BITSAT 2021
A biconvex lens has a radius of curvature of magnitude $20 \mathrm{~cm}$. Which one of the following options best describe the image formed of an object of height $2 \mathrm{~cm}$ placed $30 \mathrm{~cm}$ from the lens?

Options:
A Virtual, upright, height $=1 \mathrm{~cm}$
B Virtual, upright, height $=0.5 \mathrm{~cm}$
C Real, inverted, height $=4 \mathrm{~cm}$
D Real, inverted, height $=1 \mathrm{~cm}$

The correct answer is:
Real, inverted, height $=4 \mathrm{~cm}$

Solution:
In the branch containing \(L\) and \(R_{2}\)

\(\begin{aligned} &\mathrm{i}=\frac{\mathrm{E}}{\mathrm{R}_{2}}\left(1-\mathrm{e}^{-\frac{\mathrm{R}_{2} \mathrm{t}}{\mathrm{L}}}\right) \\ &\frac{\mathrm{di}}{\mathrm{dt}}=\frac{\mathrm{E}}{\mathrm{R}_{2}} \mathrm{e}^{-\frac{\mathrm{R}_{2} \mathrm{t}}{\mathrm{L}}} \cdot \frac{\mathrm{R}_{2}}{\mathrm{~L}}=\frac{\mathrm{E}}{\mathrm{L}} \mathrm{e}^{-\frac{\mathrm{R}_{2} \mathrm{t}}{\mathrm{L}}} \\ &\therefore \mathrm{V}_{\mathrm{L}}=\mathrm{L} \frac{\mathrm{di}}{\mathrm{dt}}=\mathrm{Ee}^{-\frac{\mathrm{R}_{2} \mathrm{t}}{\mathrm{L}}} \\ &=12 \mathrm{e}^{-\frac{2 \mathrm{t}}{400 \times 10^{-3}}}=12 \mathrm{e}^{-5 \mathrm{t}} \mathrm{V} \end{aligned}\)

Topic: Physics, Ray Optics, BITSAT, BITSAT 2016
A bicycle tyre is filled with air having pressure of $270 \mathrm{kPa}$ at $27^{\circ} \mathrm{C}$. The approximate pressure of the air in the tyre when the temperature increases to $36^{\circ} \mathrm{C}$ is

Options:
A $270 \mathrm{kPa}$
B $262 \mathrm{KPa}$
C $278 \mathrm{kPa}$
D $360 \mathrm{kPa}$

The correct answer is:
$278 \mathrm{kPa}$

Solution:
From the ideal gas equation $\mathrm{PV}=\mathrm{nRT}$ Here, volume is constant $\therefore \frac{\mathrm{P}_1}{\mathrm{~T}_1}=\frac{\mathrm{P}_2}{\mathrm{~T}_2}$
Here, $T_1=27+273=300 \mathrm{~K}$
$P_1=270 \mathrm{kPa}$
$\mathrm{T}_2=36+273=309 \mathrm{~K}$
$\Rightarrow P_2=\frac{P_1}{T_1} \times T_2=\frac{270 \times(309)}{300}=278 \mathrm{kPa}$

Topic: Physics, Kinetic Theory of Gases, BITSAT, BITSAT 2023 (Memory Based Paper 1)
A black die and a white die are rolled. Find the probability that the number shown by the black die will be more than twice that shown by the white die.

Options:
A \(1 / 8\)
B \(1 / 6\)
C \(1 / 3\)
D \(1 / 4\)

The correct answer is:
\(1 / 6\)

Solution:
The number of favourable cases are shown below:
\(\begin{array}{cl}\text { Number on white die } & \text { Number on black die } \\ 1 & 3 \\ 1 & 4 \\ 1 & 5 \\ 1 & 6 \\ 2 & 5 \\ 2 & 6\end{array}\)
There are 6 favourable cases in which the number on black die is more than twice the number on the white die.
\(\begin{aligned}
& \therefore \mathrm{m}=6 \\
& \mathrm{n}=\text { Total number of cases }=6 \times 6
\end{aligned}\)
( \(\therefore\) with each die there are six possibilities)
\(\therefore\) Probability \(\mathrm{p}=\frac{\mathrm{m}}{\mathrm{n}}=\frac{6}{6 \times 6}=\frac{1}{6}\)

Topic: Mathematics, Functions, BITSAT, BITSAT 2010
A block A of mass 100 kg rests on another block B of mass 200 kg and is tied to a wall as shown in the figure. The coefficient of friction between A and B is 0.2 and that between B and ground is 0.3. The minimum force required to move the block B is g=10 m s-2

Options:
A 900 N
B 200 N
C 1100 N
D 700 N

The correct answer is:
1100 N

Solution:
Friction force between A and B and between block B and surface S will oppose F

∴ F=FAB+FBS=μABMAg+μBSmA+mBg

=0.2×100×10+0.3100+200×10

=200+900=1100 N

Topic: Physics, Laws of Motion, BITSAT, BITSAT 2017
A block of mass m is placed on a smooth inclined wedge $A B C$ of inclination $\theta$ as shown in the figure. The wedge is given an acceleration 'a' towards the right. The relation between a and $\theta$ for the block to remain stationary on the wedge is

Options:
A $a=\frac{g}{\operatorname{cosec} \theta}$
B $a=\frac{g}{\sin \theta}$
C $a=g \tan \theta$
D $a=g \cos \theta$

The correct answer is:
$a=g \tan \theta$

Solution:
Let the mass of block is m. It will remains stationary if forces acting on it are in equilibrium. i.e., $m a \cos \theta=m g \sin \theta$

$\Rightarrow a=g \tan \theta$

Here $\mathrm{ma}=$ Pseudo force on block,

$\mathrm{mg}=$ weight

Topic: Physics, Laws of Motion, BITSAT, BITSAT 2013
A block of mass $\mathrm{M}$ is pulled along a horizontal frictionless surface by a rope of mass $\mathrm{M} / 2$. If a force $2 \mathrm{Mg}$ is applied at one end of the rope, the force which the rope exerts on the block is -

Options:
A $2 \mathrm{Mg} / 3$
B $2 \mathrm{Mg}$
C $4 \mathrm{Mg} / 3$
D zero

The correct answer is:
$4 \mathrm{Mg} / 3$

Solution:
No Solution available

Topic: Physics, Laws of Motion, BITSAT, BITSAT 2009
A block of wood floats in water with (4/5)th of its volume submerged. If the same block just floate in a liquid, the density of liquid in kg m-3 is

Options:
A 1250
B 600
C 400
D 800

The correct answer is:
800

Solution:
Let V be the volume of the block. When block floats in water, then Vblock g=45vρwater g

or Pblock=45ρwater .....i

When block floats in liquid.

Vρblock g=Vρliquidg

ρblock =ρliquid

ρliquid =45ρwater [from Eq. (i)]

=45×103 kg m-3

=800 kg m-3

Topic: Physics, Mechanical Properties of Fluids, BITSAT, BITSAT 2018
A bob is hanging over a pulley inside a car through, a string. The second end of the string is in the hand of a person standing in the car. The car is moving with constant acceleration 'a' directed horizontallyas shown in figure. Other end of the string is pulled with constant acceleration 'a' vertically. The tension in the string is equal to -

Options:
A $m \sqrt{g^{2}+a^{2}}$
B $m \sqrt{g^{2}+a^{2}}-m a$
C $\mathrm{m} \sqrt{\mathrm{g}^{2}+\mathrm{a}^{2}}+\mathrm{ma}$
D $\mathrm{m}(\mathrm{g}+\mathrm{a})$

The correct answer is:
$\mathrm{m} \sqrt{\mathrm{g}^{2}+\mathrm{a}^{2}}+\mathrm{ma}$

Solution:
Applying Newton's law

along string

$$

\Rightarrow \mathrm{T}-\mathrm{m} \sqrt{\mathrm{g}^{2}+\mathrm{a}^{2}}=\mathrm{ma}

$$

or $\mathrm{T}=\mathrm{m} \sqrt{\mathrm{g}^{2}+\mathrm{a}^{2}}+\mathrm{ma}$

Topic: Physics, Laws of Motion, BITSAT, BITSAT 2013
A body is moving unidirectionally under the influence of a source of constant power. The square of its displacement in time t is proportional to

Options:
A t3
B t2
C t
D t

The correct answer is:
t3

Solution:
By work-energy theorem,

W=P×t=12mv2

⇒v2=2Ptm

v=2Ptm1/2

or dsdt=2Ptm1/2

∫ds=∫2Ptm1/2dt

s=2Pm1/2·23t3/2

i.e. s∝t3/2

s2∝t3

Topic: Physics, Work Power Energy, BITSAT, BITSAT 2019
A body is projected, making an acute angle with the horizontal. If angle between velocity $\vec{v}$ and acceleration $\overrightarrow{\mathrm{g}}$ is $\theta,$ then

Options:
A $\theta=90^{\circ}$
B $\quad \theta=0^{\circ}$
C $90^{\circ}<\theta<0^{\circ}$
D $0^{\circ}<\theta<180^{\circ}$

The correct answer is:
$0^{\circ}<\theta<180^{\circ}$

Solution:
Here velocity is acting upwards when projectile is going upwards and acceleration is downwards. The angle $\theta$ between $\overrightarrow{\mathrm{v}}$ and $\overrightarrow{\text { a }}$ is more than $0^{\circ}$ and less than $180^{\circ}$

Topic: Physics, Motion In Two Dimensions, BITSAT, BITSAT 2013
A body is thrown vertically upwards from A, the top of the tower, reaches the ground in time $t_{1}$. If it is thrown vertically downwards from A with the same speed, it reaches the ground in time $\mathrm{t}_{2}$. If it is allowed to fall freely from $\mathrm{A},$ then the time it takes to reach the ground is given by

Options:
A $\mathrm{t}=\frac{\mathrm{t}_{1}+\mathrm{t}_{2}}{2}$
B $\mathrm{t}=\frac{\mathrm{t}_{1}-\mathrm{t}_{2}}{2}$
C $\quad \mathrm{t}=\sqrt{\mathrm{t}_{1} \mathrm{t}_{2}}$
D $\mathrm{t}=\sqrt{\frac{\mathrm{t}_{1}}{\mathrm{t}_{2}}}$

The correct answer is:
$\quad \mathrm{t}=\sqrt{\mathrm{t}_{1} \mathrm{t}_{2}}$

Solution:
Let the body is projected vertically upwards from A with a speed $u_{0}$ Using equation, $\mathrm{s}=\mathrm{ut}+\frac{1}{2} \mathrm{at}^{2}$

For case $(1)-\mathrm{h}=\mathrm{u}_{0} \mathrm{t}_{1}-\frac{1}{2} \mathrm{gt}_{1}^{2} \quad \ldots(1)$

For case (2) $-\mathrm{h}=\mathrm{u}_{0} \mathrm{t}_{2}-\frac{1}{2} \mathrm{gt}_{1}^{2}$

Subtracting eq (2) from (1), we get

$$

\begin{array}{l}

0=\mathrm{u}_{0}\left(\mathrm{t}_{2}+\mathrm{t}_{1}\right)+\frac{1}{2} \mathrm{~g}\left(\mathrm{t}_{2}^{2}-\mathrm{t}_{1}^{2}\right) \\

\Rightarrow \mathrm{u}_{0}=\frac{1}{2} \mathrm{~g}\left(\mathrm{t}_{1}-\mathrm{t}_{2}\right)

\end{array}

$$

Putting the value of $\mathrm{u}_{0}$ in eq (2), we get

$$

-\mathrm{h}=-\left(\frac{1}{2}\right) \mathrm{g}\left(\mathrm{t}_{1}-\mathrm{t}_{2}\right) \mathrm{t}_{2}-\left(\frac{1}{2}\right) \mathrm{gt}_{2}^{2}

$$

$$

\Rightarrow \mathrm{h}=\frac{1}{2} \mathrm{~g}\left(\mathrm{t}_{1} \mathrm{t}_{2}\right)

$$

For case $3, \mathrm{u}_{0}=0, \mathrm{t}=?$

$$

-\mathrm{h}=0 \times \mathrm{t}-\left(\frac{1}{2}\right) \mathrm{gt}^{2}

$$

$\mathrm{h}=\left(\frac{1}{2}\right) \mathrm{gt}^{2}$

Comparing eq. (4) and $(5),$ we get

$$

\frac{1}{2} \mathrm{gt}^{2}=\frac{1}{2} \mathrm{gt}_{1} \mathrm{t}_{2} \quad \therefore \mathrm{t}=\sqrt{\mathrm{t}_{1} \mathrm{t}_{2}}

$$

Topic: Physics, Motion In One Dimension, BITSAT, BITSAT 2015
A body is travelling in a circle at a constant speed. It

Options:
A has a constant velocity
B is not accelerated
C has an inward radial acceleration
D has an outward radial acceleration

The correct answer is:
has an inward radial acceleration

Solution:
Body moves with constant speed it means that tangential acceleration $\mathrm{a}_{\mathrm{T}}=0 \& \mathrm{conly}$ centripetal acceleration $\mathrm{a}_{\mathrm{C}}$ exists whose direction is always towards the centre or inward (along the radius of the circle).

Topic: Physics, Motion In Two Dimensions, BITSAT, BITSAT 2010
A body moves with uniform acceleration, then which of the following graph is correct?

Options:
A
B
C
D

The correct answer is:

Solution:
An object is said to be moving with a uniform acceleration, if its velocity changes by equal amount in equal intervals of time. The velocity-time graph of uniformly accelerated motion is a straight line inclined to time axis. Acceleration of an object in a uniformly accelerated motion in one dimension is equal to the slope of the velocity-time graph with time axis.

Topic: Physics, Motion In One Dimension, JEE Main
A body moves with uniform acceleration, then which of the following graph is correct?

Options:
A
B
C
D

The correct answer is:

Solution:
An object is said to be moving with a uniform acceleration, if its velocity changes by equal amount in equal intervals of time. The velocity-time graph of uniformly accelerated motion is a straight line inclined to time axis. Acceleration of an object in a uniformly accelerated motion in one dimension is equal to the slope of the velocity-time graph with time axis.

Topic: Physics, Motion In One Dimension, BITSAT, BITSAT 2012
A body of density $\rho^{\prime}$ is dropped from rest at a height h into a lake of density $\rho$ where $\rho>\rho^{\prime}$ neglecting all dissipative forces, calculate the maximum depth to which the body sinks before returning to float on the surface:

Options:
A $\frac{\mathrm{h}}{\rho-\rho^{\prime}}$
B $\frac{\mathrm{h} \rho^{\prime}}{\rho}$
C $\frac{\mathrm{h} \rho^{\prime}}{\rho-\rho^{\prime}}$
D $\frac{h \rho}{\rho-\rho^{\prime}}$

The correct answer is:
$\frac{\mathrm{h} \rho^{\prime}}{\rho-\rho^{\prime}}$

Solution:
The effective acceleration of the body

$\mathrm{g}^{\prime}=\left(1-\frac{\rho}{\rho^{\prime}}\right) \mathrm{g}$

Now, the depth to which the body sinks

$\mathrm{h}^{\prime}=\left(\frac{\mathrm{u}^{2}}{2 \mathrm{~g}^{\prime}}\right)=\frac{2 \mathrm{gh}}{2 \mathrm{~g}^{\prime}}=\frac{\mathrm{gh}}{\mathrm{g}^{\prime}}=\frac{\mathrm{h} \times \rho^{\prime}}{\rho-\rho^{\prime}}$

Topic: Physics, Mechanical Properties of Fluids, BITSAT, BITSAT 2020
A body of length $1 \mathrm{~m}$ having cross-sectional area $0.75 \mathrm{~m}^{2}$ has heat flow through it at the rate of 6000 Joule/sec. Then find the temperature difference if $\mathrm{K}=200 \mathrm{Jm}^{-1} \mathrm{~K}^{-1}$

Options:
A $20^{\circ} \mathrm{C}$
B $40^{\circ} \mathrm{C}$
C $80^{\circ} \mathrm{C}$
D $100^{\circ} \mathrm{C}$

The correct answer is:
$40^{\circ} \mathrm{C}$

Solution:
$\frac{Q}{t}=\frac{\operatorname{KA} \Delta \theta}{\ell} \Rightarrow 6000=\frac{200 \times 0.75 \times \Delta \theta}{1}$
$\therefore \Delta \theta=\frac{6000 \times 1}{200 \times 0.75}=40^{\circ} \mathrm{C}$

Topic: Physics, Thermal Properties of Matter, BITSAT, BITSAT 2009
A body of mass 1 kg is executing simple harmonic motion. Its displacement ycm at t seconds is given by y=6sin100t+π4. Its maximum kinetic energy is

Options:
A 6 J
B 18 J
C 24 J
D 36 J

The correct answer is:
18 J

Solution:
Given, a=6 cm=6×10-2 m

ω=100 rad s-1,m=1 kg

Kmax=12mω2a2

=12×1×1002×6×10-22

=18 J

Topic: Physics, Oscillations, BITSAT, BITSAT 2017
A body of mass $2 \mathrm{~kg}$ is placed on a horizontal surface having kinetic friction 0.4 and static friction $0.5 .$ If the force applied on the body is $2.5 \mathrm{~N},$ then the frictional force acting on the body will be $\left[\mathrm{g}=10 \mathrm{~ms}^{-2}\right]$

Options:
A $8 \mathrm{~N}$
B $10 \mathrm{~N}$
C $20 \mathrm{~N}$
D $2.5 \mathrm{~N}$

The correct answer is:
$2.5 \mathrm{~N}$

Solution:
Limiting friction $=0.5 \times 2 \times 10=10 \mathrm{~N}$

The applied force is less than force of friction, therefore the force of friction is equal to the applied force.

Topic: Physics, Laws of Motion, BITSAT, BITSAT 2020
A body of mass m is moving in a straight line with momentum p. Starting at time t=0, a force F=at acts in the same direction on the moving particle during time interval of T. So that its momentum changes from p to 2p. The value of T is

Options:
A 2pa
B pa
C 22pa
D 2pa

The correct answer is:
2pa

Solution:
Given, F=at

when, t=0, then linear momentum =p

when, t=T, then linear momentum =2p

According to Newton's law of motion, applied force, F=dpdt

or dp=Fdt

dp=atdt

or ∫p2pdp=a∫0Ttdt

2p-p=a·T22

⇒2pa=T2⇒T=2pa

Topic: Physics, Laws of Motion, BITSAT, BITSAT 2019
A body of weight $2 \mathrm{~kg}$ is suspended as shown in the figure. The tension $\mathrm{T}_{1}$ in the horizontal string (in $\mathrm{kg}$ wt ) is

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

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

Solution:
The tension in the above string can be divided into two components where the component opposite to weight will balance the weight while the component of this tension opposite to \(\mathrm{T}_{1}\) will be equal to \(\mathrm{T}_{1}\).

\(\mathrm{T} \sin \theta=2 \mathrm{~kg} \mathrm{wt}\)

\(\mathrm{T} \cos \theta=\mathrm{T}_{1}\)

The first equation gives

\(\mathrm{T}=4 \mathrm{~kg} \mathrm{wt}\)

So \(\mathrm{T}_{1}\) will be,

\(\mathrm{T}_{1}=2 \sqrt{3} \mathrm{~kg} \mathrm{wt}\)

Topic: Physics, Laws of Motion, BITSAT, BITSAT 2021
$\mathrm{A} \sigma$ bonded molecule $\mathrm{MX}_{3}$ is T-shaped. The number of non bonded pair of electrons is

Options:
A 0
B 2
C 1
D can be predicted only if atomic number is known

The correct answer is:
2

Solution:
For T-shape geometry the molecule must have 3 bonded pair and 2 lone pair of electrons.

Topic: Chemistry, Chemical Bonding and Molecular Structure, BITSAT, BITSAT 2011
A bowl filled with very hot soup cools from $98^{\circ} \mathrm{C}$ to $86^{\circ} \mathrm{C}$ in 2 minutes when the room temperature is $22^{\circ} \mathrm{C}$. How long it will take to cool from $75^{\circ} \mathrm{C}$ to $69^{\circ} \mathrm{C}$ ?

Options:
A
2 minutes
B
1.4 minutes
C
0.5 minute
D
1 minute

The correct answer is:
1.4 minutes

Solution:
The rate of heat flow is given by
$\begin{aligned}
& \frac{\Delta Q}{\Delta t}=-K\left(T-T_0\right) \Rightarrow \frac{\Delta Q}{\Delta t}=-K\left(T_{a v g}-T_0\right) \\
& \text { Initially } T_{a v g}=\frac{T_1+T_2}{2} \\
& \text { Here, } T_1=98^{\circ} \mathrm{C} \\
& T_2=86^{\circ} \mathrm{C}
\end{aligned}$
$\begin{aligned} & \text { (i) } \frac{\mathrm{ms} \times 12}{2}=-\mathrm{K}\left(\frac{98+86}{2}-22\right) \\ & 6=-\frac{\mathrm{K}}{\mathrm{ms}}\left[\frac{98+86}{2}-22\right] \quad(\because \Delta \mathrm{Q}=\mathrm{ms} \Delta \mathrm{T})\end{aligned}$
Now, cool from $\mathrm{T}_1^{\prime}=75^{\circ} \mathrm{C}$
$\begin{aligned}
& \mathrm{T}_2=69^{\circ} \mathrm{C} \\
& \mathrm{T}_{\mathrm{avg}}=\frac{\mathrm{T}_1^{\prime}+\mathrm{T}_2^{\prime}}{2}=\frac{75+69}{2}
\end{aligned}$
(ii) $\frac{\mathrm{ms} \times 6}{\Delta \mathrm{t}}=-\mathrm{K}\left(\frac{75+69}{2}-22\right)$
Divide equation (ii) by (i), we have
$\begin{aligned}
& \frac{6}{\Delta t(6)}=\frac{50}{70} \\
& \Delta t=\frac{7}{5}=1.4 \mathrm{~min}
\end{aligned}$

Topic: Physics, Thermal Properties of Matter, BITSAT, BITSAT 2023 (Memory Based Paper 2)