datavorous/entrance-exam-dataset huggingface dataset:
A transverse wave propagating on a stretched string of linear density 3×10-4 kgm-1 is represented by the equation

y=0.2sin(1.5x+60t)

where, x in metres and t is in seconds. The tension in the string (in Newton) is

Options:
A 0.24
B 0.48
C 1.20
D 1.80

The correct answer is:
0.48

Solution:
Equation of wave,

y=0.2sin[1.5x+60t]

Comparing with standard equation, we get

y=Asin[kx+ωt]

k=1.5=2πλ

and ω=60=2πT

∴ Velocity of wave, v=ωk

=601.5=40 m s-1

Velocity of wave in stretched string, v=Tm

where, m is the linear density, T is tension in the string.

So, T=v2m=(40)2×3×10-4

=0.48 N

Topic: Physics, Waves and Sound, BITSAT, BITSAT 2017
A trolley having mass of 200 kg moves with uniform speed of 36 kmh-1 on a frictionless track. A child of mass 20 kg runs on the trolley from one end to the other (10 m away) with a speed of 4 m s-1 relative to the trolley in a direction opposite to its inotion and ultimately jumps out of the trolley. With how much velocity has the trolley moved from the time the child begins to run?

Options:
A 10.36 m s-1
B 11.36 m s-1
C 12.36 m s-1
D 14.40 m s-1

The correct answer is:
10.36 m s-1

Solution:
since, no external force is acting on the system, we can apply conservation of linear momentum.

Speed of 200 kg trolley =36×100060×60=10 m s-1

If u be the initial velocity of trolley, vb be the absolute velocity of the boy after the beginning of journey of the boy, their relative velocity is 4.

So, v'-vb=4⇒vb=v'-4

Now, applying law of conservation of momentum,

Momentum before the boy begins tu run
= Momentum after the beginning of boy's running

⇒ 220×10=200v'+20v'-4

2200=220v'-80⇒220v'=2280

v'=2280220=10.36 m s-1

Topic: Physics, Center of Mass Momentum and Collision, BITSAT, BITSAT 2018
A tunnel has been dug through the centre of the earth and a ball is released in it. It executes S.H.M. with time period

Options:
A 42 minutes
B 1 day
C 1 hour
D 84.6 minutes

The correct answer is:
84.6 minutes

Solution:
$\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{R}}{\mathrm{g}}}=2 \pi \sqrt{\frac{64 \times 10^{6}}{9.8}}$

$=2 \times \frac{22}{7} \times \frac{8 \times 10^{3}}{7 \times \sqrt{2}}$

$=\frac{\sqrt{2} \times 22 \times 8 \times 1000}{49 \times 60} \min =84.6 \mathrm{~min}$

Topic: Physics, Oscillations, BITSAT, BITSAT 2014
A U - shaped tube contains a liquid of density rho and it is rotated about the left dotted line as shown in the figure. Find the difference in the levels of liquid column.

Options:
A $\frac{\omega^{2} \mathrm{~L}^{2}}{2 g}$
B $\frac{\omega^{2} L^{2}}{2 \sqrt{2} g}$
C $\frac{2 \omega^{2} \mathrm{~L}^{2}}{g}$
D $\frac{2 \sqrt{2} \omega^{2} \mathrm{~L}^{2}}{g}$

The correct answer is:
$\frac{\omega^{2} \mathrm{~L}^{2}}{2 g}$

Solution:
Weight of liquid of height \(\mathrm{H}=\frac{\pi \mathrm{d}^{2}}{4} \times \mathrm{H} \rho \times \mathrm{g} \rightarrow\) (i)

Let us consider a mass dm situated at a distance \(x\) from \(A\) as shown in the figure. The centripetal force required for the mass to rotate \(=(\mathrm{dm}) \omega^{2}\)

\(\therefore\) The total centripetal force required for the mass of length \(L\) to rotate \(=\int_{0}^{1}\left(\rho \mathrm{x} \frac{\pi \mathrm{d}^{2}}{4}\right) \mathrm{xdx} \times\left(\mathrm{x} \omega^{2}\right)=\rho \times \frac{\pi \mathrm{d}^{2}}{4} \times \omega^{2} \times \frac{1^{2}}{2} \rightarrow\) (ii)

This centripetal force required is provided by the weight of liquid of height 'H' From (i) and (ii)

\(\begin{aligned} & \frac{\pi \mathrm{d}^{2}}{4} \times \mathrm{H} \times \rho \times \mathrm{g}=\rho \times \frac{\pi \mathrm{d}^{2}}{4} \times \omega^{2} \times \frac{1^{2}}{2} \\ & \Rightarrow \mathrm{H}=\frac{\omega^{2} \mathrm{l}^{2}}{2 \mathrm{~g}} \end{aligned}\)

Topic: Physics, Mechanical Properties of Fluids, BITSAT, BITSAT 2021
A uniform cube of side a and mass m rests on a rough horizontal table. A horizontal force $\mathrm{F}$ is applied normal to one of the faces at a point that is directly above the centre of the face, at a height $3 \mathrm{a} / 4$ above the base. The minimum value of $\mathrm{F}$ for which the cube begins to topple an edge is (assume that cube does not slide)

Options:
A $\frac{\mathrm{mg}}{3}$
B $\frac{\mathrm{mg}}{2}$
C $\frac{2 \mathrm{mg}}{3}$
D $\frac{3 \mathrm{mg}}{4}$

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

Solution:
For toppling about edge $x$ x At the moment of toppling the normal force pass through axis $\mathrm{xx}$ '.
$$
\mathrm{F}_{\min } \frac{3 \mathrm{a}}{4}=\mathrm{mg} \frac{\mathrm{a}}{2} \text { or } \mathrm{F}_{\min }=\frac{2 \mathrm{mg}}{3}
$$

Topic: Physics, Rotational Motion, JEE Main
A uniform cube of side a and mass m rests on a rough horizontal table. A horizontal force $\mathrm{F}$ is applied normal to one of the faces at a point that is directly above the centre of the face, at a height $3 \mathrm{a} / 4$ above the base. The minimum value of $\mathrm{F}$ for which the cube begins to topple an edge is (assume that cube does not slide)

Options:
A $\frac{\mathrm{mg}}{3}$
B $\frac{\mathrm{mg}}{2}$
C $\frac{2 \mathrm{mg}}{3}$
D $\frac{3 \mathrm{mg}}{4}$

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

Solution:
For toppling about edge $x$ x At the moment of toppling the normal force pass through axis $\mathrm{xx}$ '.
$$
\mathrm{F}_{\min } \frac{3 \mathrm{a}}{4}=\mathrm{mg} \frac{\mathrm{a}}{2} \text { or } \mathrm{F}_{\min }=\frac{2 \mathrm{mg}}{3}
$$

Topic: Physics, Rotational Motion, BITSAT, BITSAT 2009
A uniform electric field and uniform magnetic field are acting along the same direction in a certain region. If an electron is projected in the region such that its velocity is pointed along the direction of fields, then the electron

Options:
A will turn towards right of direction of motion
B speed will decrease
C speed will increase
D will turn towards left direction of motion

The correct answer is:
speed will decrease

Solution:
$\overrightarrow{\mathrm{v}}$ and $\overrightarrow{\mathrm{B}}$ are in same direction so that magnetic force on electron becomes zero, only electric force acts. But force on electron due to electric field is opposite to the direction of velocity.

Topic: Physics, Magnetic Effects of Current, BITSAT, BITSAT 2013
A uniform rod of length l and mass m is free to rotate in a vertical plane about A. The rod initially in horizontal position is released. The initial angular acceleration of the rod is (Moment of inertia of rod about A is ml23)

Options:
A 3g2l
B 23g
C 3g2l2
D mgl2

The correct answer is:
3g2l

Solution:
The moment of inertia of the uniform rod about an axis through one end and perpendicular to its length is

=ml23

where, m is the mass and l its length.

Torque τ=Iα acting on centre of gravity of rod is given by

τ=mgl2

also, ml23·α=mgl2

α=3g2l

Topic: Physics, Rotational Motion, JEE Main
A uniform rod of length l and mass m is free to rotate in a vertical plane about A. The rod initially in horizontal position is released. The initial angular acceleration of the rod is (Moment of inertia of rod about A is ml23)

Options:
A 3g2l
B 23g
C 3g2l2
D mgl2

The correct answer is:
3g2l

Solution:
The moment of inertia of the uniform rod about an axis through one end and perpendicular to its length is

=ml23

where, m is the mass and l its length.

Torque τ=Iα acting on centre of gravity of rod is given by

τ=mgl2

also, ml23·α=mgl2

α=3g2l

Topic: Physics, Rotational Motion, BITSAT, BITSAT 2017
A uniform solid cylindrical roller of mass m is being pulled on horizontal surface with force F parallel to the surface applied at its centre. If the acceleration of the cylinder is a and it is rolling without slipping, then the value of F is

Options:
A 2 ma
B 3 ma
C 3 ma2
D 5 ma2

The correct answer is:
3 ma2

Solution:
By Newton's law of motion

F-fs=ma ...(i)

and torque, τ=lα

where, α= tangential component of acceleration
I= moment of inertia

⇒ fR=lα

fR=MR22·aR∴I=MR22and ,α=aR

⇒ f=ma2

∴ F=f+ma

=ma2+ma=3ma2

Topic: Physics, Rotational Motion, BITSAT, BITSAT 2019
A vector perpendicular to the plane containing the vectors $\hat{i}-2 \hat{j}-\hat{k}$ and $3 \hat{i}-2 \hat{j}-\hat{k}$ is inclined to the vector $\hat{i}+\hat{j}+\hat{k}$ at an angle

Options:
A $\tan ^{-1} \sqrt{14}$
B $\sec ^{-1} \sqrt{14}$
C $\tan ^{-1} \sqrt{15}$
D
None of these

The correct answer is:
$\tan ^{-1} \sqrt{14}$

Solution:
A vector perpendicular to the plane is
$\begin{aligned}
& (\hat{i}-2 \hat{j}-\hat{k}) \times(3 \hat{i}-2 \hat{j}-\hat{k}) \\
& =\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
1 & -2 & -1 \\
3 & -2 & -1
\end{array}\right|=-2 \hat{j}+4 \hat{k} \\
& \Rightarrow \text { unit vector } \hat{a}=\frac{-2 \hat{j}+4 \hat{k}}{\sqrt{4+16}}=\frac{-2 \hat{j}+4 \hat{k}}{2 \sqrt{5}}
\end{aligned}$
Angle between the unit vector and $\vec{r}=\hat{i}+\hat{j}+\hat{k}$
$=\cos ^{-1} \frac{\vec{r} \cdot \hat{a}}{|\vec{r}| \cdot|\hat{a}|}=\cos ^{-1} \frac{1}{\sqrt{15}}=\tan ^{-1} \sqrt{14}$

Topic: Mathematics, Vector Algebra, BITSAT, BITSAT 2023 (Memory Based Paper 2)
A vessel containing 1 more of $\mathrm{O}_{2}$ gas (molar mass 32) at temperature $\mathrm{T}$. The pressure of the gas is p. An identical vessel containing one mole of he gas (molar mass 4 ) at temperature $2 \mathrm{~T}$ has a pressure of

Options:
A $\mathrm{p} / 8$
B $\mathrm{p}$
C $2 \mathrm{p}$
D $8 \mathrm{p}$

The correct answer is:
$2 \mathrm{p}$

Solution:
Applying gas equation, $\mathrm{pV}=\mathrm{nRT}$ We can write, $\mathrm{p}_{1} \mathrm{~V}=\mathrm{n}_{1} \mathrm{RT}_{1}$ and $\mathrm{p}_{2} \mathrm{~V}=\mathrm{n}_{2} \mathrm{RT}_{2}$

$\Rightarrow \quad \frac{\mathrm{p}_{2}}{\mathrm{p}_{1}}=\frac{\mathrm{n}_{2}}{\mathrm{n}_{1}} \times \frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}=\frac{1}{1} \times \frac{2 \mathrm{~T}}{\mathrm{~T}}=2$

$\Rightarrow \mathrm{p}_{2}=2 \mathrm{p}$

Topic: Physics, Thermodynamics, BITSAT, BITSAT 2013
A vessel of depth $2 \mathrm{~d} \mathrm{~cm}$ is half filled with a liquid of refractive index $\mu_{1}$ and the upper half with a liquid of refractive index $\mu_{2}$. The apparent depth of the vessel seen perpendicularly is

Options:
A $\left(\frac{\mu_{1} \mu_{2}}{\mu_{1}+\mu_{2}}\right) \mathrm{d}$
B $\left(\frac{1}{\mu_{1}}+\frac{1}{\mu_{2}}\right) \mathrm{d}$
C $\left(\frac{1}{\mu_{1}}+\frac{1}{\mu_{2}}\right) 2 \mathrm{~d}$
D $\left(\frac{1}{\mu_{1} \mu_{2}}\right) 2 \mathrm{~d}$

The correct answer is:
$\left(\frac{1}{\mu_{1}}+\frac{1}{\mu_{2}}\right) \mathrm{d}$

Solution:
Apparent depth $=\mathrm{d} / \mu_{1}+\mathrm{d} / \mu_{2}$

Topic: Physics, Ray Optics, BITSAT, BITSAT 2020
A vessel of depth $2 \mathrm{~d} \mathrm{~cm}$. is halffilled with a liquid of refractive index $\mu_{1}$ and the upper half with a liquid of refractive index $\mu_{2} .$ The apparent depth of the vessel seen perpendicularly is -

Options:
A $d\left(\frac{\mu_{1} \mu_{2}}{\mu_{1}+\mu_{2}}\right)$
B $d\left(\frac{1}{\mu_{1}}+\frac{1}{\mu_{2}}\right)$
C $2 d\left(\frac{1}{\mu_{1}}+\frac{1}{\mu_{2}}\right)$
D $2 d\left(\frac{1}{\mu_{1} \mu_{2}}\right)$

The correct answer is:
$d\left(\frac{1}{\mu_{1}}+\frac{1}{\mu_{2}}\right)$

Solution:
$\mathrm{h}^{\prime}=\frac{\mathrm{d}_{1}}{\mu_{1}}+\frac{\mathrm{d}_{2}}{\mu_{2}}=\mathrm{d}\left(\frac{1}{\mu_{1}}+\frac{1}{\mu_{2}}\right)$

Topic: Physics, Ray Optics, BITSAT, BITSAT 2011
A vessel of volume 20L contains a mixture of hydrogen and helium at temperature of $27^{\circ} \mathrm{C}$ and pressure 2 atm. The mass of mixture is $5 \mathrm{~g}$. Assuming the gases to be ideal, the ratio of mass of hydrogen to that of helium in the given mixture will be

Options:
A 1: 2
B 2: 3
C 2: 1
D 2: 5

The correct answer is:
2: 5

Solution:
Let there are $\mathrm{n}_{1}$ moles of hydrogen and $\mathrm{n}_{2}$ moles of helium in the given mixture. As $\mathrm{Pv}$ $=\mathrm{nRT}$

Then the pressure of the mixture

$$

\begin{array}{r}

P=\frac{\mathrm{n}_{1} \mathrm{RT}}{\mathrm{V}}+\frac{\mathrm{n}_{2} \mathrm{RT}}{\mathrm{V}}=\left(\mathrm{n}_{1}+\mathrm{n}_{2}\right) \frac{\mathrm{RT}}{\mathrm{V}} \\

2 \times 101.3 \times 10^{3}=\left(\mathrm{n}_{1}+\mathrm{n}_{2}\right) \times \frac{(8.3 \times 300)}{20 \times 10^{-3}}

\end{array}

$$

or,

$$

\begin{array}{l}

\left(\mathrm{n}_{1}+\mathrm{n}_{2}\right)=\frac{2 \times 101.3 \times 10^{3} \times 20 \times 10^{-3}}{(8.3)(300)} \\

\text { or, } \mathrm{n}_{1}+\mathrm{n}_{2}=1.62

\end{array}

$$

The mass of the mixture is (in grams)

$$

\begin{array}{l}

\mathrm{n}_{1} \times 2+\mathrm{n}_{2} \times 4=5 \\

\Rightarrow \quad\left(\mathrm{n}_{1}+2 \mathrm{n}_{2}\right)=2.5 \quad \ldots

\end{array}

$$

Solving the eqns. (1) and (2), we get $\mathrm{n}_{1}=0.74$ and $\mathrm{n}_{2}=0.88$

Hence, $\frac{\mathrm{m}_{\mathrm{H}}}{\mathrm{m}_{\mathrm{He}}}=\frac{0.74 \times 2}{0.88 \times 4}=\frac{1.48}{3.52}=\frac{2}{5}$

Topic: Physics, Thermodynamics, BITSAT, BITSAT 2015
A vessel with water is placed on a weighing pan and it reads $600 \mathrm{~g}$. Now a ball of mass $40 \mathrm{~g}$ and density $0.80 \mathrm{~g} \mathrm{~cm}^{-3}$ is sunk into the water with a pin of negligible volume, as shown in figure keeping it sunk. The weighing pan will show a reading

Options:
A $600 \mathrm{~g}$
B $550 \mathrm{~g}$
C $650 \mathrm{~g}$
D $632 \mathrm{~g}$

The correct answer is:
$650 \mathrm{~g}$

Solution:
Volume of ball $=\frac{40}{0.8}=50 \mathrm{~cm}^{3}$

Downthrust on water $=50 \mathrm{~g}$. Therefore reading is $650 \mathrm{~g}$.

Topic: Physics, Mechanical Properties of Fluids, BITSAT, BITSAT 2020
A wave $y=a \sin (\omega t-k x)$ on a string meets with another wave producing a node at $\mathrm{x}=0 .$ Then the equation of the unknown wave is

Options:
A $y=a \sin (\omega t+k x)$
B $y=-a \sin (\omega t+k x)$
C $y=a \sin (\omega t-k x)$
D $y=-a \sin (\omega t-k x)$

The correct answer is:
$y=-a \sin (\omega t+k x)$

Solution:
Equation of a wave $y_{1}=a \sin (\omega t-k x)$....(i)
Let equations of another wave may be, $y_{2}=a \sin (\omega t+k x)$....(ii)
$y_{3}=-a \sin (\omega t+k x) \quad \ldots .$ (iii)
If Eq. (i) propagate with Eq. (ii), we get $\mathrm{y}=2 \mathrm{a} \cos \mathrm{kx} \sin \omega \mathrm{t}$
If Eq. (i), propagate with Eq. (iii), we get $y=-2 a \sin k x \cos \omega t$
At $x=0, y=0,$ wave produce node
So, Eq.(iii) is the equation of unknown wave

Topic: Physics, Waves and Sound, BITSAT, BITSAT 2010
A wheel is rotating at $900 \mathrm{r}$.p.m. about its axis. When power is cut offit comes to rest in 1 minute. The angular retardation in $\mathrm{rad} / \mathrm{s}^{2}$ is

Options:
A $\pi / 2$
B $\pi / 4$
C $\pi / 6$
D $\pi / 8$

The correct answer is:
$\pi / 2$

Solution:
Angular retardation,

$\begin{aligned} \alpha &=\frac{\omega_{2}-\omega_{1}}{t}=\frac{2 \pi\left(\mathrm{n}_{2}-\mathrm{n}_{1}\right)}{\mathrm{t}} \\ &=\frac{2 \pi(0-900 / 60)}{60}=\frac{\pi}{2} \mathrm{rad} / \mathrm{s}^{2} \end{aligned}$

Topic: Physics, Motion In Two Dimensions, BITSAT, BITSAT 2020
A wheel of radius $\mathrm{R}$ rolls on the ground with a uniform velocity $\mathrm{v}$. The relative acceleration of topmost point of the wheel with respect to the bottom most point is

Options:
A $\frac{\mathrm{v}^{2}}{\mathrm{R}}$
B $\frac{2 \mathrm{v}^{2}}{\mathrm{R}}$
C $\frac{\mathrm{v}^{2}}{2 \mathrm{R}}$
D $\frac{4 \mathrm{v}^{2}}{\mathrm{R}}$

The correct answer is:
$\frac{2 \mathrm{v}^{2}}{\mathrm{R}}$

Solution:
As $a_{\mathrm{CM}}=0 \quad\left[\mathrm{v}_{\mathrm{CM}}=\right.$ constant $]$, Tangential acceleration of each point $\left|\overrightarrow{\mathrm{a}}_{\mathrm{AB}}\right|=\frac{2 \mathrm{v}^{2}}{\mathrm{R}}$

Topic: Physics, Motion In Two Dimensions, BITSAT, BITSAT 2012
A whistle of frequency 500Hz tied to the end of a string of length 1.2 m revolves at 400 rev min-1. A listener standing some distance away in the plane of rotation of whistle hears frequencies in the range. (Speed of sound =340 m s-1)

Options:
A 436 to 574
B 426 to 586
C 426 to 574
D

436 to 586

The correct answer is:

436 to 586

Solution:
vs=rω=1.2×2πf ∵ω=2πf

=1.2×2×3.14×40060=50 m s-1

vmin=vv+vs·v

=340340+50×500

=436 Hz

vmax=vv-vs·v

=340340-50×500

=586 Hz

Topic: Physics, Waves and Sound, BITSAT, BITSAT 2017
A whistle of frequency $1000 \mathrm{~Hz}$ is sounded on a car travelling towards a cliff with velocity of 18 $\mathrm{~m} \mathrm{~s}^{-1}$ normal to the cliff. If $\mathrm{c}=330 \mathrm{~m} \mathrm{~s}^{-1}$, then the apparent frequency of the echo as heard by the car driver is nearly

Options:
A $1115 \mathrm{~Hz}$
B $115 \mathrm{~Hz}$
C $67 \mathrm{~Hz}$
D $47.2 \mathrm{~Hz}$

The correct answer is:
$1115 \mathrm{~Hz}$

Solution:
By the concept of accoustic, the observer and source are moving towards each other, each with a velocity of $18 \mathrm{~m} \mathrm{~s}^{-1}$ $\therefore \mathrm{v}^{\prime}=\frac{330+18}{330-18} \times 1000 \approx 1115 \mathrm{~Hz}$

Topic: Physics, Waves and Sound, BITSAT, BITSAT 2020
A wholesale merchant wants to start the business of cereal with \(₹ 24000\). Wheat is \(₹ 400\) per quintal and rice is \(₹ 600\) per quintal. He has capacity to store 200 quintal cereal. He earns the profit \(₹ 25\) per quintal on wheat and ₹ 40 per quintal on rice. If he store \(\mathrm{x}\) quintal rice and y quintal wheat, then for maximum profit the objective function is

Options:
A \(25 x+40 y\)
B \(40 x+25 y\)
C \(400 \mathrm{x}+600 \mathrm{y}\)
D \(\frac{400}{40} \mathrm{x}+\frac{600}{25} \mathrm{y}\)

The correct answer is:
\(40 x+25 y\)

Solution:
For maximum profit, \(z=40 x+25 y\).

Topic: Mathematics, Sequences and Series, BITSAT, BITSAT 2010
A wire $34 \mathrm{~cm}$ long is to be bent in the form of a quadrilateral of which each angle is $90^{\circ} .$ What is the maximum area which can be enclosed inside the quadrilateral?

Options:
A $68 \mathrm{~cm}^{2}$
B $70 \mathrm{~cm}^{2}$
C $71.25 \mathrm{~cm}^{2}$
D $72.25 \mathrm{~cm}^{2}$

The correct answer is:
$72.25 \mathrm{~cm}^{2}$

Solution:
Let one side of quadrilateral be $\mathrm{x}$ and another side be $y$ so, $2(x+y)=34$

or $(x+y)=17$

We know from the basic principle that for a given perimeter square has the maximum area, $\mathrm{so}, \mathrm{x}=\mathrm{y}$ and putting this value in equation (i)

$\mathrm{x}=\mathrm{y}=\frac{17}{2}$

Area $=x \cdot y=\frac{17}{2} \times \frac{17}{2}=\frac{289}{4}=72.25$

Topic: Mathematics, Area Under Curves, BITSAT, BITSAT 2016
A wire carrying current I has the shape as shown in adjoining figure. Linear parts of the wire are very long and parallel to $\mathrm{X}$ -axis while semicircular portion of radius $\mathrm{R}$ is lying in Y-Z plane. Magnetic field at point $\mathrm{O}$ is:

Options:
A $\overrightarrow{\mathrm{B}}=-\frac{\mu_{0}}{4 \pi} \frac{\mathrm{I}}{\mathrm{R}}(\mu \hat{\mathrm{i}} \times 2 \hat{\mathrm{k}})$
B $\quad \overrightarrow{\mathrm{B}}=-\frac{\mu_{0}}{4 \pi} \frac{\mathrm{I}}{\mathrm{R}}(\pi \hat{\mathrm{i}}+2 \hat{\mathrm{k}})$
C $\overrightarrow{\mathrm{B}}=\frac{\mu_{0}}{4 \pi} \frac{\mathrm{I}}{\mathrm{R}}(\pi \hat{\mathrm{i}}-2 \hat{\mathrm{k}})$
D $\overrightarrow{\mathrm{B}}=\frac{\mu_{0}}{4 \pi} \frac{\mathrm{I}}{\mathrm{R}}(\pi \hat{\mathrm{i}}+2 \hat{\mathrm{k}})$

The correct answer is:
$\quad \overrightarrow{\mathrm{B}}=-\frac{\mu_{0}}{4 \pi} \frac{\mathrm{I}}{\mathrm{R}}(\pi \hat{\mathrm{i}}+2 \hat{\mathrm{k}})$

Solution:
No Solution

Topic: Physics, Magnetic Effects of Current, BITSAT, BITSAT 2016
A wire of a certain material is stretched slowly by ten per cent. Its new resistance and specific resistance become respectively:

Options:
A $1.2$ times, $1.3$ times
B $1.21$ times, same
C both remain the same
D $1.1$ times, $1.1$ times

The correct answer is:
$1.21$ times, same

Solution:
\(\mathrm{R}=\frac{\rho \mathrm{l}}{\mathrm{A}}\)

Now, \(1=1+\frac{1}{10}=\frac{111}{10}\)

and therefore, \(\mathrm{A}=\frac{10 \mathrm{~A}}{11}\)

So \(\mathrm{R}^{\prime}=\frac{\mathrm{p} \times\left(\frac{111}{10}\right)}{\frac{10 \mathrm{~A}}{11}}=\frac{\rho \mathrm{l}}{\mathrm{A}} \times \frac{(11)^{2}}{(10)^{2}}=1.21 \mathrm{R}\)

Now resistance becomes \(1.21\) times of initial and specific resistance is the intrinsic property so remains same.

Topic: Physics, Current Electricity, BITSAT, BITSAT 2021
A wire $X$ is half the diameter and half the length of a wire $Y$ of similar material. The ratio of resistance of $\mathrm{X}$ to that of $\mathrm{Y}$ is

Options:
A 8: 1
B 4: 1
C 2: 1
D 1: 1

The correct answer is:
2: 1

Solution:
$\mathrm{R} \propto \frac{\ell}{\mathrm{D}^{2}} \Rightarrow \frac{\mathrm{Rx}}{\mathrm{Ry}}=\frac{2}{1}$

Topic: Physics, Current Electricity, BITSAT, BITSAT 2014
A zener diode of voltage $\mathrm{V}_{\mathrm{Z}}(=6 \mathrm{~V})$ is used to maintain a constant voltage across a load resistance $\mathrm{R}_{\mathrm{L}}(=1000 \Omega)$ by using a series resistance $R_{\mathrm{s}}(=100 \Omega)$. If the e.m.f. of source is $\mathrm{E}(=9 \mathrm{~V})$, what is the power being dissipated in Zener diode?

Options:
A $0.144$ watt
B $0.324$ watt
C $0.244$ watt
D $0.544$ watt

The correct answer is:
$0.144$ watt

Solution:
Here, $E=9 \mathrm{~V} ; \mathrm{V}_{\mathrm{z}}=6 ; \mathrm{R}_{\mathrm{L}}=1000 \Omega$ and $\mathrm{R}_{\mathrm{s}}=100 \Omega$,

Potential drop across series resistor $\mathrm{V}=\mathrm{E}-\mathrm{V}_{\mathrm{Z}}=9-6=3 \mathrm{~V}$

Current through series resistance $R_{S}$ is $\mathrm{I}=\frac{\mathrm{V}}{\mathrm{R}}=\frac{3}{100}=0.03 \mathrm{~A}$

Current through load resistance $R_{L}$ is $\mathrm{I}_{\mathrm{L}}=\frac{\mathrm{V}_{\mathrm{Z}}}{\mathrm{R}_{\mathrm{L}}}=\frac{6}{1000}=0.006 \mathrm{~A}$

Current through Zener diode is $\mathrm{I}_{\mathrm{Z}}=\mathrm{I}-\mathrm{I}_{\mathrm{L}}=0.03-0.006=0.024 \mathrm{amp} .$ Power dissipated in Zener diode is $\mathrm{P}_{\mathrm{Z}}=\mathrm{V}_{\mathrm{Z}} \mathrm{I}_{\mathrm{Z}}=6 \times 0.024=0.144 \mathrm{Watt}$

Topic: Physics, Semiconductors, BITSAT, BITSAT 2021
$\mathrm{ABC}$ is a triangular park with $\mathrm{AB}=\mathrm{AC}=100 \mathrm{~m} . \mathrm{A}$ TV tower stands at the mid-point of $\mathrm{BC}$. The angles of elevation of the top of the tower at A, B, C are $45^{\circ}, 60^{\circ}, 60^{\circ}$ respectively. The height of the tower is

Options:
A $50 \mathrm{~m}$
B $50 \sqrt{3} \mathrm{~m}$
C $50 \sqrt{2} \mathrm{~m}$
D $50(3-\sqrt{3}) \mathrm{m}$

The correct answer is:
$50 \sqrt{3} \mathrm{~m}$

Solution:
$\tan 45^{\circ}=\frac{\mathrm{PQ}}{\mathrm{AQ}}=\frac{\mathrm{h}}{\mathrm{AQ}} \Rightarrow \mathrm{h}=\mathrm{AQ}$

Where $\mathrm{PQ}$ is tower and $\mathrm{ABC}$ is the park, with $\mathrm{Q}$ being mid point of the side $\mathrm{BC}$ and $\mathrm{PQ}=\mathrm{h}$ Also, $A Q^{2}+B Q^{2}=100^{2}$

$\Rightarrow \mathrm{h}^{2}+\mathrm{h}^{2} \cot ^{2} 60^{\circ}=100^{2}$

$\Rightarrow \mathrm{h}^{2}\left[1+\frac{1}{3}\right]=100^{2}$

$\Rightarrow \mathrm{h}^{2}=\frac{3 \times 100^{2}}{4} \Rightarrow \mathrm{h}=50 \sqrt{3}$

Topic: Mathematics, Heights and Distances, BITSAT, BITSAT 2020
According to IUPAC system, the correct name of the compound having the formula

Options:
A 2 -ethyl-3-methyl but- 1 -ene
B 2 -methyl pent- 3 -ene
C 2 -methyl but- 1 -ene
D None of these

The correct answer is:
None of these

Solution:

Topic: Chemistry, General Organic Chemistry, BITSAT, BITSAT 2010
Acetaldehyde and acetone can be distinguished by :

Options:
A Iodoform test
B Nitroprusside test
C Fehlings solution test
D $C$ \& P test

The correct answer is:
Fehlings solution test

Solution:
Acetaldehyde is easily oxidised to acetic acid by a mild oxidising agent like Fehling solution. Acetone is not easily oxidised. Both acetone and acetaldehyde give iodoform test. Other two conditions are not relevant to aldehydes and ketones.

Topic: Chemistry, Aldehydes and Ketones, BITSAT, BITSAT 2011
Acyclicprocess ABCD is shown in P-V diagram for an ideal gas. Which of the following diagram represents the same process?

Options:
A
B
C
D

The correct answer is:

Solution:
$\mathrm{AB}$ : Isobaric expansion

$\mathrm{BC}:$ Isothermal expansion

CD : Isochoric

DA: Isothermal compression

Topic: Chemistry, Thermodynamics (C), JEE Main
Acyclicprocess ABCD is shown in P-V diagram for an ideal gas. Which of the following diagram represents the same process?

Options:
A
B
C
D

The correct answer is:

Solution:
$\mathrm{AB}$ : Isobaric expansion

$\mathrm{BC}:$ Isothermal expansion

CD : Isochoric

DA: Isothermal compression

Topic: Chemistry, Thermodynamics (C), BITSAT, BITSAT 2020