datavorous/entrance-exam-dataset huggingface dataset:
In a radioactive material the activity at time $t_1$, is $A_1$ and at a later time $t_2$, it is $A_2$. If the decay constant of the material is $\lambda$, then

Options:
A
$A_1=A_2 e^{-\lambda\left(t_1-t_2\right)}$

B
$A_1=A_2 e^{\lambda\left(t_1-t_2\right)}$

C
$A_1=A_2\left(t_2 / t_1\right)$

D
$A_1=A_2$

The correct answer is:

$A_1=A_2 e^{-\lambda\left(t_1-t_2\right)}$

Solution:
From radioactive decay law,
$$
-\frac{d N}{d t} \propto N \text { or }-\frac{d N}{d t}=\lambda N
$$
Thus,
$$
A=-\frac{d N}{d t} \text { or } A=\lambda N
$$
$$
\Rightarrow \quad A=\lambda N_0 e^{-\lambda t}
$$
Where, $A_0=\lambda N_0$ is the activity of the radioactive material at time, $t=0$
At time, $t_1 \quad A_1=A_0 e^{-\lambda t_1}$
At time, $t_2 \quad A_2=A_0 e^{-\lambda t_2}$
On dividing Eq. (ii) by Eq. (iii), we have
$$
\begin{aligned}
\quad \frac{A_1}{A_2} & =\frac{e^{-\lambda t_1}}{e^{-\lambda t_2}}=e^{-\lambda\left(t_1-t_2\right)} \\
\Rightarrow \quad \quad A_1 & =A_2 e^{-\lambda\left(t_1-t_2\right)}
\end{aligned}
$$

Topic: Physics, Nuclear Physics, BITSAT, BITSAT 2022
In a reflecting telescope, a secondary mirror is used to:

Options:
A reduce the problem of mechanical support
B remove spherical aberration
C make chromatic aberration zero
D move the eyepiece outside the telescopic tube

The correct answer is:
move the eyepiece outside the telescopic tube

Solution:
(d) To redirect the light that enters the telescope to the eyepiece or camera. The primary mirror of a reflecting telescopes gathers the light and reflects towards the secondary mirror which then reflect the light towards the eyepiece allowing the observer to see image.
It has advantage of a large focal length in a short telescope.

Topic: Physics, Ray Optics, BITSAT, BITSAT 2023 (Memory Based Paper 1)
In a region of space having a uniform electric field $\mathrm{E}$, a hemispherical bowl of radius $\mathrm{r}$ is placed. The electric flux $\phi$ through the bowl is

Options:
A $2 \pi \mathrm{RE}$
B $4 \pi R^{2} E$
C $2 \pi R^{2} E$
D $\pi R^{2} E$

The correct answer is:
$2 \pi R^{2} E$

Solution:
$\phi=\mathrm{E}(\mathrm{ds}) \cos \theta=\mathrm{E}\left(2 \pi \mathrm{r}^{2}\right) \cos 0^{\circ}=2 \pi \mathrm{r}^{2} \mathrm{E}$

Topic: Physics, Electrostatics, BITSAT, BITSAT 2012
In a region, the potential is represented by $\mathrm{V}(\mathrm{x}, \mathrm{y}, \mathrm{z})=6 \mathrm{x}-8 \mathrm{xy}-8 \mathrm{y}+6 \mathrm{yz},$ where $\mathrm{V}$ is in volts and $\mathrm{x}, \mathrm{y}, \mathrm{z}$ are in metres. The electric force experienced by a charge of 2 coulomb situated at point (1,1,1) is :

Options:
A $6 \sqrt{5} \mathrm{~N}$
B $30 \mathrm{~N}$
C $24 \mathrm{~N}$
D $4 \sqrt{35} \mathrm{~N}$

The correct answer is:
$4 \sqrt{35} \mathrm{~N}$

Solution:
$\overrightarrow{\mathrm{E}}=-\frac{\partial \mathrm{V}}{\partial \mathrm{x}} \hat{\mathrm{i}}-\frac{\partial \mathrm{V}}{\partial \mathrm{y}} \hat{\mathrm{j}}-\frac{\partial \mathrm{V}}{\partial \mathrm{z}} \hat{\mathrm{k}}$
$=-[(6-8 \mathrm{y}) \hat{\mathrm{i}}+(-8 \mathrm{x}-8+6 \mathrm{z}) \hat{\mathrm{j}}+(6 \mathrm{y}) \hat{\mathrm{k}}]$
$\mathrm{At}(1,1,1), \overrightarrow{\mathrm{E}}=2 \hat{\mathrm{i}}+10 \hat{\mathrm{j}}-6 \hat{\mathrm{k}}$
$\Rightarrow \quad(\overrightarrow{\mathrm{E}})=\sqrt{2^{2}+10^{2}+6^{2}}=\sqrt{140}=2 \sqrt{35}$
$\therefore \quad \mathrm{F}=\mathrm{q} \overrightarrow{\mathrm{E}}=2 \times 2 \sqrt{35}=4 \sqrt{35}$

Topic: Physics, Electrostatics, BITSAT, BITSAT 2010
In a resonance column first and second resonance are obtained at depths $24 \mathrm{~cm}$ and $78 \mathrm{~cm}$ the third resonance will be obtained at depth.

Options:
A
$160 \mathrm{~cm}$

B
$132 \mathrm{~cm}$

C
$131 \mathrm{~cm}$

D
$152 \mathrm{~cm}$

The correct answer is:

$132 \mathrm{~cm}$

Solution:
Given, first resonance column at depth, $l_1=24 \mathrm{~cm}$
Second resonance column at depth, $l_2=78 \mathrm{~cm}$
Third resonance column at depths, $l_3=$ ?
As
$$
\begin{aligned}
& l_1+x=\frac{\lambda}{4}=24 \quad \quad \ldots (i)\\
& l_2+x=\frac{3 \lambda}{4}=78 \quad \quad \ldots (ii)\\
& l_3+x=\frac{5 \lambda}{4} \quad \quad \ldots (iii)
\end{aligned}
$$
From Eqs. (i) and (ii),
$$
x=\frac{l_2-3 l_1}{2}=\frac{78-3(24)}{2}=3 \quad \quad \ldots (iv)
$$
From Eqs. (iii) and (i), $\frac{l_3+x}{l_1+x}=5$
$$
l_3=5 l_1+5 x-x=5 l_1+4 x
$$
Substituting $l_1$ and $x$ value, we get
$$
l_3=5(24)+4(3) \Rightarrow l_3=132 \mathrm{~cm} .
$$

Topic: Physics, Oscillations, BITSAT, BITSAT 2022
In a $\mathrm{S}_{\mathrm{N}} 2$ substitution reaction of the type $\mathrm{R}-\mathrm{Br}+\mathrm{Cl}^{-} \stackrel{\mathrm{DMF}}{\longrightarrow} \mathrm{R}-\mathrm{Cl}+\mathrm{Br}^{-}$

which one of the following has the highest relative rate?

Options:
A $\mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{CH}_{2} \mathrm{Br}$
B
C
D $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{Br}$

The correct answer is:
$\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{Br}$

Solution:
$\mathrm{S}_{\mathrm{N}} 2$ mechanism is followed in case of primary and secondary halides i.e., $\mathrm{S}_{\mathrm{N}}^{2}$ reaction is favoured by small groups on the carbon atom attached to halogens so

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main
In a $\mathrm{S}_{\mathrm{N}} 2$ substitution reaction of the type $\mathrm{R}-\mathrm{Br}+\mathrm{Cl}^{-} \stackrel{\mathrm{DMF}}{\longrightarrow} \mathrm{R}-\mathrm{Cl}+\mathrm{Br}^{-}$

which one of the following has the highest relative rate?

Options:
A $\mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{CH}_{2} \mathrm{Br}$
B
C
D $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{Br}$

The correct answer is:
$\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{Br}$

Solution:
$\mathrm{S}_{\mathrm{N}} 2$ mechanism is followed in case of primary and secondary halides i.e., $\mathrm{S}_{\mathrm{N}}^{2}$ reaction is favoured by small groups on the carbon atom attached to halogens so

Topic: Chemistry, Haloalkanes and Haloarenes, BITSAT, BITSAT 2013
In a semiconductor diode, the barrier potential offers opposition to

Options:
A holes in P-region only
B free electrons in N-region only
C majority carriers in both regions
D majority as well as minority carriers in both regions

The correct answer is:
majority carriers in both regions

Solution:
No Solution available

Topic: Physics, Semiconductors, BITSAT, BITSAT 2011
In a sequence of $21$ terms, the first $11$ terms are in $AP$ with common difference $2$ and the last $11$ terms are in $GP$ with common ratio $2$. If the middle term of $AP$ be equal to the middle term of the $GP$, then the middle term of the entire sequence is

Options:
A
$-\frac{10}{31}$

B
$\frac{10}{31}$

C
$\frac{32}{31}$

D
$-\frac{31}{32}$

The correct answer is:

$-\frac{10}{31}$

Solution:
Since, the first $11$ terms are in $AP$, $d=2$
$\therefore \quad a_{11}=a+10 d=a+20$
The middle term of $AP$ is
$T_6=a+5 d=a+10$
For the next $11$ terms in $GP$
$r=2$
$\therefore$ The middle term of $GP$ is $b(2)^5$ where, $b$ is the first term of a $GP$ which is the last term of $AP$
$b(2)^5=(a+20) 32$
According to the given condition,
$\begin{aligned}
\Rightarrow \quad a+10 & =(a+20) 32 \\
\Rightarrow \quad 31 a & =10-640 \\
a & =-\frac{630}{31}
\end{aligned}$
$\therefore$ Middle term of entire sequence is $11th$ term
$\begin{aligned} \therefore \quad T_{11} & =\frac{-630}{31}+10 \times d \\ & =\frac{-630}{31}+10 \times 2=\frac{-10}{31}\end{aligned}$

Topic: Mathematics, Sequences and Series, BITSAT, BITSAT 2022
In a simple pendulum of length I the bob is pulled aside from its equilibrium position through an angle $\theta$ and then released. The bob passes through the equilibrium position with speed

Options:
A $\sqrt{2 g l(1+\cos \theta)}$
B $\sqrt{2 g l \sin \theta}$
C $\sqrt{2 \mathrm{~g} l}$
D $\sqrt{2 \mathrm{~g} l(1-\cos \theta)}$

The correct answer is:
$\sqrt{2 \mathrm{~g} l(1-\cos \theta)}$

Solution:
If $l$ is length of pendulum and $\theta$ be angular amplitude then from principle of conservation of energy

(Initial total energy) $0+\mathrm{mgh}=\frac{1}{2} m v^{2}+0$ (Final total energy) $\Rightarrow v=\sqrt{2 g h}=\sqrt{2 g l(1-\cos \theta)}$

Topic: Physics, Oscillations, BITSAT, BITSAT 2021
In a solid lattice, the cation has left a lattice site and is located at an interstitial position, the lattice defect is :

Options:
A
Interstitial defect
B
Valency defect
C
Frenkel defect
D
Schottky defect

The correct answer is:
Frenkel defect

Solution:
Frenkel defect is due to dislocation of ion from its usual lattice site to interstitial position.

Topic: Chemistry, Solid State, BITSAT, BITSAT 2023 (Memory Based Paper 2)
In a statistical investigation of 1003 families of Calcutta, it was found that 63 families has neither a radio nor a T.V, 794 families has a radio and 187 has T.V. The number of families in that group having both a radio and a T.V is

Options:
A 36
B 41
C 32
D None of these

The correct answer is:
41

Solution:
Let $\mathrm{R}$ be the set of families having a radio and $T$ the set families having a T.V. then $\mathrm{n}(\mathrm{R} \cup \mathrm{T})=$ The number of families having at least on of the radio and $\mathrm{T} \mathrm{V}=1003-63=940$ $n(\mathrm{R})=794$ and $\mathrm{n}(\mathrm{T})=187$

Let $x$ families have both a radio and a T.V. Then number of families who have only radio $=$ $794-x$

And the number of families who have only T.V. $=187-x$

Erom Venn diagram, $794-x+x-187-x=940$ $\Rightarrow 981-x=940$ or $x=981-940=41$

Hence, the required number of families having both a radio and a T.V.=41

Topic: Mathematics, Sets and Relations, BITSAT, BITSAT 2020
In a test an examiner either guesses or copies or knows the answer to a multiple choice question with 4 choices. The probability that he/she makes a guess is 13. The probability that he/she copies the answer is 16. If the probability that the answer is correct, given that he/she copied it is 18, then the probability that he/she knows the answer to a question given that he/she correctly answered it, is

Options:
A 2729
B 2629
C 2529
D 2429

The correct answer is:
2429

Solution:
Let A be the event that the examinee gives the correct answer.

Let G, C, K stand for guessing, copying and knowing. respectively.

Given, PC=16, PG=13

and PAC=18

⇒PC+PG+PK=1

⇒16+13+PK=1

∴ PK=12

Also, PAK=1, for if the examinee knows, he/she will correctly answer it and PAG=14,

since there are four choices.

Now, total probability PA=PGPAG+PCPAC+PKPAK

=13·14+16·18+12·1

=112+148+12=4+1+2448=2948

Now, PKA=PKPAKPA

=12·12948=12·4829=2429

Topic: Mathematics, Probability, BITSAT, BITSAT 2018
In a thermodynamic process, the pressure of a fixed mass of a gas is changed in such a manner that the gas releases $20 \mathrm{~J}$ of heat and $8 \mathrm{~J}$ of work is done on the gas. If the initial internal energy of the gas was $30 \mathrm{~J}$, then the final internal energy will be

Options:
A $2 \mathrm{~J}$
B $42 \mathrm{~J}$
C $18 \mathrm{~J}$
D $58 \mathrm{~J}$

The correct answer is:
$18 \mathrm{~J}$

Solution:
Given $\Delta \mathrm{Q}=-20 \mathrm{~J}, \mathrm{~W}=-8 \mathrm{~J}$

Using Ist law $\Delta \mathrm{Q}=\Delta \mathrm{U}+\Delta \mathrm{W}$

$$

\begin{array}{l}

\Rightarrow \Delta Q=-20+8=-12 \mathrm{~J} \\

\mathrm{Uf}=-12+30=18 \mathrm{~J}

\end{array}

$$

Topic: Physics, Thermodynamics, BITSAT, BITSAT 2014
In a triangle $\mathrm{ABC},$ if $\mathrm{a}=2, \mathrm{~B}=60^{\circ}$ and $\mathrm{C}=75^{\circ}$, then b equals

Options:
A $\sqrt{3}$
B $\sqrt{6}$
C $\sqrt{9}$
D $1+\sqrt{2}$

The correct answer is:
$\sqrt{6}$

Solution:
$\mathrm{A}=180^{\circ}-60^{\circ}-75^{\circ}=180^{\circ}-135^{\circ}=45^{\circ}$
Now, $\frac{\mathrm{a}}{\sin \mathrm{A}}=\frac{\mathrm{b}}{\sin \mathrm{B}}$
$\Rightarrow \frac{2}{\sin 45^{\circ}}=\frac{\mathrm{b}}{\sin 60^{\circ}} \Rightarrow \mathrm{b}=\frac{2 \cdot(\sqrt{3} / 2)}{1 / \sqrt{2}}=\sqrt{6}$

Topic: Mathematics, Properties of Triangles, BITSAT, BITSAT 2012
In a triangle \(\mathrm{ABC}, \angle \mathrm{C}=90^{\circ}\), then \(\frac{a^2-b^2}{a^2+b^2}\) is equal to :

Options:
A \(\sin (\mathrm{A}+\mathrm{B})\)
B \(\sin (A-B)\)
C \(\cos (\mathrm{A}+\mathrm{B})\)
D \(\sin \left(\frac{\mathrm{A}-\mathrm{B}}{2}\right)\)

The correct answer is:
\(\sin (A-B)\)

Solution:
\(\begin{aligned}
& \mathrm{A}+\mathrm{B}=180^{\circ}-\mathrm{C}=90^{\circ} \\
& \mathrm{a}=2 \mathrm{R} \sin \mathrm{A}, b=2 \mathrm{R} \sin \mathrm{B}, c=2 \mathrm{R} \sin \mathrm{C} \\
& \begin{array}{l}
\therefore \frac{a^2-b^2}{a^2+b^2}=\frac{\sin ^2 \mathrm{~A}-\sin ^2 \mathrm{~B}}{\sin ^2 \mathrm{~A}+\sin ^2 \mathrm{~B}} \\
\quad=\frac{\sin (\mathrm{A}+\mathrm{B}) \sin (\mathrm{A}-\mathrm{B})}{\sin ^2 \mathrm{~A}+\sin ^2\left(90^{\circ}-\mathrm{A}\right)} \\
{\left[\because \mathrm{A}+\mathrm{B}=90^{\circ}\right]} \\
=\frac{\sin 90^{\circ} \sin (\mathrm{A}-\mathrm{B})}{\sin ^2 \mathrm{~A}+\cos ^2 \mathrm{~A}}=\sin (\mathrm{A}-\mathrm{B})
\end{array}
\end{aligned}\)

Topic: Mathematics, Heights and Distances, BITSAT, BITSAT 2009
In a YDSE, the light of wavelength $\lambda=5000 Ã$ is used, which emerges in phase from two slits a distance $\mathrm{d}=3 \times 10^{-7} \mathrm{~m}$ apart. Atransparent sheet of thickness $\mathrm{t}=1.5 \times 10^{-7} \mathrm{~m}$ refractive index $\mu=1.17$ is placed over one of the slits. what is the new angular position of the central maxima of the interference pattern, from the centre of the screen? Find the value of $y$.

Options:
A $4.9^{\circ}$ and $\frac{D(\mu-1) t}{2 d}$
B $4.9^{\circ}$ and $\frac{\mathrm{D}(\mu-1) \mathrm{t}}{\mathrm{d}}$
C $3.9^{\circ}$ and $\frac{D(\mu+1) t}{d}$
D $2.9^{\circ}$ and $\frac{2 \mathrm{D}(\mu+1) \mathrm{t}}{\mathrm{d}}$

The correct answer is:
$4.9^{\circ}$ and $\frac{\mathrm{D}(\mu-1) \mathrm{t}}{\mathrm{d}}$

Solution:
The path difference when transparent sheet is introduced $\Delta x=(\mu-1) t$

If the central maxima occupies position of nth fringe, then $(\mu-1) \mathrm{t}=\mathrm{n} \lambda=\mathrm{d} \sin \theta$

$$

\Rightarrow \sin \theta=\frac{(\mu-1) \mathrm{t}}{\mathrm{d}}=\frac{(1.17-1) \times 1.5 \times 10^{-7}}{3 \times 10^{-7}}=0.085

$$

Therefore, angular position of central maxima

$$

\theta=\sin ^{-1}(0.085)=4.88^{\circ} \approx 4.9

$$

For small angles, $\sin \theta=\theta=\tan \theta$

$$

\Rightarrow \quad \tan \theta=\frac{y}{D}

$$

$\therefore \quad \frac{\mathrm{y}}{\mathrm{D}}=\frac{(\mu-1) \mathrm{t}}{\mathrm{d}} \Rightarrow \mathrm{y}=\frac{\mathrm{D}(\mu-1) \mathrm{t}}{\mathrm{d}}$

Topic: Physics, Wave Optics, BITSAT, BITSAT 2015
In a young's double slit arrangement frings are produced using light of wavelength $4000 Å$. One slit is covered by a thin plate of glass of refractive index $1.4$ and the other with another glass plate of same thickness but of refractive index 1.7. By doing so the central bright shifts to original sixth fringe from centre. Thickness of glass plate is

Options:
A
$2 \mu \mathrm{m}$

B
$8 \mu \mathrm{m}$

C
$11 \mu \mathrm{m}$

D
$16 \mu \mathrm{m}$

The correct answer is:

$8 \mu \mathrm{m}$

Solution:
As, shift $\Delta x=\frac{\beta}{\lambda}(\mu-1) t$

Shift due to one plate $\Delta x_1=\frac{\beta}{\lambda}\left(\mu_1-1\right) t$
Shift due to another path $\Delta x_2=\frac{\beta}{\lambda}\left(\mu_2-1\right) t$
Net shift $\Delta x=\Delta x_2-\Delta x_1=\frac{\beta}{\lambda}\left(\mu_2-\mu_1\right) t$
Also, it is given that $\Delta x=6 \beta$
$$
\begin{aligned}
& \text { Hence, } 6 \beta=\frac{\beta}{\lambda}\left(\mu_1-\mu_2\right) t \\
& \Rightarrow \quad t=\frac{6 \lambda}{\left(\mu_1-\mu_2\right)}=\frac{6 \times 4000}{(1.7-1.4)}-80000 Å \\
& \Rightarrow \quad t=8 \times 10^{-6}=8 \mu \mathrm{m}
\end{aligned}
$$

Topic: Physics, Wave Optics, BITSAT, BITSAT 2022
In a Young's double slit experiment the angular width of a fringe formed on a distant screen is $1^{\circ}$. The wavelength fo the light used is $6280 \AA$. What is the distance between the two coherent sources?

Options:
A $0.036 \mathrm{~mm}$
B $0.12 \mathrm{~mm}$
C $6 \mathrm{~mm}$
D $4 \mathrm{~mm}$

The correct answer is:
$0.036 \mathrm{~mm}$

Solution:
The angular fringe width is given by $\alpha=\frac{\lambda}{\mathrm{d}}$
where $\lambda$ is wavelength and $\mathrm{d}$ is the distance between two coherent sources. Thus
$\begin{array}{l}
\mathrm{d}=\frac{\lambda}{\alpha} \\
\text { Given, } \lambda=6280 \AA, \alpha=1^{\circ}=\frac{\pi}{180} \mathrm{radian}
\end{array}$
Thus $\mathrm{d}=\frac{6280 \times 10^{-10}}{3.14} \times 180$
$=3.6 \times 10^{-5} \mathrm{~m}=0.036 \mathrm{~mm}$

Topic: Physics, Wave Optics, BITSAT, BITSAT 2012
In a Young's double slit experiment, the intensities at two points, for the path difference $\frac{\lambda}{4}$ and $\frac{\lambda}{3}$ ( $\lambda$ being the wavelength of light used) are $I_1$ and $I_2$ respectively. If $I_0$ denotes the intensity produced by each one of the individual slits, then $\frac{\mathrm{I}_1+\mathrm{I}_2}{\mathrm{I}_0}=\ldots \ldots$

Options:
A 3
B 5
C 7
D 10

The correct answer is:
3

Solution:
Resultant intensity in Young's double slit experiment
$\mathrm{I}=4 \mathrm{I}_0 \cos ^2\left(\frac{\Delta \phi}{2}\right)$
For path difference $\frac{\lambda}{4}$ phase difference,
$\Delta \phi=\frac{2 \pi}{\lambda} \times \frac{\lambda}{4}=\frac{\pi}{4}$
$\therefore \mathrm{I}_1=4 \mathrm{I}_0 \cos ^2\left(\frac{\pi}{4}\right)=2 \mathrm{I}_0$
For path difference $\frac{\lambda}{3}$
$I_2=4 I_0 \cos ^2\left(\frac{2 \pi}{\lambda} \times \frac{\lambda}{3}\right)=I_0$
$\therefore \frac{\mathrm{I}_1+\mathrm{I}_2}{\mathrm{I}_0}=3$

Topic: Physics, Wave Optics, BITSAT, BITSAT 2023 (Memory Based Paper 1)
In an adiabatic process, no transfer of heat takes place between system and surrounding. Choose the correct option for free expansion of an ideal gas under adiabatic condition from the following.

Options:
A
$q=0, \Delta T \neq 0, W=0$

B
$q \neq 0, \Delta T=0, W=0$

C
$q=0, \Delta T=0, W=0$

D
$q=0, \Delta T < 0, W \neq 0$

The correct answer is:

$q=0, \Delta T=0, W=0$

Solution:
Free expansion, $W=0$
Adiabatic process, $q=0$
$\Delta U=q+W=0$, this means that internal energy remains constant. Therefore, $\Delta T=0$ in ideal gas there is no intermolecular attraction.
Hence, when such a gas expands under adiabatic conditions into a vacuum, no heat is absorbed or evolved, since no external work is done to separate the molecules.

Topic: Chemistry, Thermodynamics (C), BITSAT, BITSAT 2022
In an adiabatic process, the pressure is increased by $\frac{2}{3} \% .$ If $\gamma=\frac{3}{2},$ then the volume decreases by nearly

Options:
A $\frac{4}{9} \%$
B $\frac{2}{3} \%$
C $1 \%$
D $\frac{9}{4} \%$

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

Solution:
$\mathrm{PV}^{3 / 2}=\mathrm{K}$

$\log \mathrm{P}+\frac{3}{2} \log \mathrm{V}=\log \mathrm{K}$

$\frac{\Delta \mathrm{P}}{\mathrm{P}}+\frac{3}{2} \frac{\Delta \mathrm{V}}{\mathrm{V}}=0$

$\frac{\Delta \mathrm{V}}{\mathrm{V}}=-\frac{2}{3} \frac{\Delta \mathrm{P}}{\mathrm{P}}$ or $\frac{\Delta \mathrm{V}}{\mathrm{V}}=\left(-\frac{2}{3}\right)\left(\frac{2}{3}\right)=-\frac{4}{9}$

Topic: Physics, Thermodynamics, BITSAT, BITSAT 2020
In an atom, an electron is moving with a speed of 600 m/s with an accuracy of 0.005%. Certainty with which the position of the electron can be located is (h=6.6×10-34 kg m2 s-1, mass of electron, me=9.1×10-31 kg)

Options:
A 1.52×10-4 m
B 5.10×10-3 m
C 1.92×10-3 m
D 3.84×10-3 m

The correct answer is:
1.92×10-3 m

Solution:
Given that velocity of electron, v=600 m/s

Accuracy of velocity =0.005%
According to Heisenberg's uncertainty principle

Δx.mΔv=h4π

∆v=0.005% or 600 m s-1

=600×0.005100=0.03

∆x×9.1×10-31×0.03=6.6×10-344×3.14

Hence, ∆x=6.6×10-344×3.14×0.03×9.1×10-31

=1.92×10-3 m

Topic: Chemistry, Structure of Atom, BITSAT, BITSAT 2018
In an equilateral triangle, the in radius, circumradius and one of the ex-radii are in the ratio

Options:
A \(2: 3: 5\)
B \(1: 2: 3\)
C \(3: 7: 9\)
D \(3: 7: 9\)

The correct answer is:
\(1: 2: 3\)

Solution:
We have \(\Delta=\frac{\sqrt{3}}{4} \mathrm{a}^2, \mathrm{~s}=\frac{3 \mathrm{a}}{2}\)
\(\therefore \quad r=\frac{\Delta}{\mathrm{s}}=\frac{\mathrm{a}}{2 \sqrt{3}}, \mathrm{R}=\frac{\mathrm{abc}}{4 \Delta}=\frac{\mathrm{a}^3}{\sqrt{3} \mathrm{a}^2}=\frac{\mathrm{a}}{\sqrt{3}}\) and \(r_1=\frac{\Delta}{s-a}=\frac{\sqrt{3} / 4 a^2}{a / 2}=\frac{\sqrt{3}}{2} a\)
Hence, \(\mathrm{r}: \mathrm{R}: \mathrm{r}_1=\frac{\mathrm{a}}{2 \sqrt{3}}: \frac{\mathrm{a}}{\sqrt{3}}: \frac{\sqrt{3}}{2} \mathrm{a}=1: 2: 3\)

Topic: Mathematics, Properties of Triangles, BITSAT, BITSAT 2011
In an experiment on photoelectric effect photons of wavelength $300 \mathrm{nm}$ eject electrons from a metal of work function $2.25 \mathrm{eV}$. A photon of energy equal to that of the most energetic electron corresponds to the following transition in the hydrogen atom:

Options:
A $n=2$ to $n=1$ state
B $n=3$ to $n=1$ state
C $n=3$ to $n=2$ state
D $n=4$ to $n=3$ state

The correct answer is:
$n=3$ to $n=2$ state

Solution:
Given $\phi=2.25 \mathrm{eV}$
$$
\lambda=300 \times 100^{-3} \mathrm{~m}
$$
Energy of the given photon,
$$
\mathrm{E}=\mathrm{hv}
$$
$$
=\frac{\mathrm{hc}}{\lambda}=\frac{663 \times 10^{-34} \times 3 \times 10^8}{300 \times 10^{-9}}=6.63 \times 10^{-19} \mathrm{~J}=4.13 \mathrm{eV}
$$
Now,
$$
\mathrm{E}=\frac{\mathrm{hc}}{\lambda}-\phi=4.137-2.25=1.88 \mathrm{eV} \rightarrow(1)
$$
So, $1.88 \mathrm{eV}$ energy is used to jump from one orbit to another orbit by electron. Therefore, energy of different orbital of hydrogen
$$
\begin{array}{rrrr}
\Rightarrow \mathrm{n}=1 & 2 & 3 & 4 \\
\mathrm{E}=-13.6 \mathrm{ev} & -3.4 & -1.51 & -0.85
\end{array}
$$
And from statement (1),
$$
1.18=E_i-E_f
$$
$\left\{E_i \& E_f \quad\right.$ are energy of initial and final orbit $\}$ And also from the above table, we can observe that,
$$
E_{\mathrm{i}}-\mathrm{E}_{\mathrm{f}}=(-1.51)-(-3.4)=1.89 \mathrm{eV}
$$
$$
=\mathrm{E}_2-\mathrm{E}_3
$$
That the electron jumps from 3rd to 2nd orbit. Therefore, with one photon of 3oonm wavelength hydrogen electron can jump from 3 to 2 .

Topic: Physics, Dual Nature of Matter, BITSAT, BITSAT 2012
In an LC oscillator, if values of inductance and capacitance become twice and eight times, respectively, then the resonant frequency of oscillator becomes $\mathrm{x}$ times its initial resonant frequency $\omega_0$. The value of $x$ is:

Options:
A $1 / 4$
B 16
C $1 / 16$
D 4

The correct answer is:
$1 / 4$

Solution:
The resonance frequency of LC oscillations circuit is
$\omega=\frac{1}{\sqrt{\mathrm{L}^{\prime} \mathrm{C}^{\prime}}} \Rightarrow \mathrm{L}^{\prime} \rightarrow 2 \mathrm{~L}$
$\mathrm{C}^{\prime} \rightarrow 8 \mathrm{C}$
$\omega=\frac{1}{\sqrt{2 \mathrm{~L} \times 8 \mathrm{C}}}=\frac{1}{4 \sqrt{\mathrm{LC}}} 9 \omega_0=\frac{1}{\sqrt{\mathrm{LC}}}$
$\omega=\frac{\omega_0}{4}$ So, $x=\frac{1}{4}$

Topic: Physics, Electromagnetic Induction, BITSAT, BITSAT 2023 (Memory Based Paper 1)
In an ore containing uranium, the ratio of $U^{238}$ to $\mathrm{Pb}^{206}$ is 3. Calculate the àge of the ore, assuming that all the lead present in the ore is the final stable product of $U^{238}$. Take the half-life of $U^{238}$ to be $4.5 \times 10^{9} \mathrm{yr}$.

Options:
A $1.6 \times 19^{3} \mathrm{yr}$
B $1.5 \times 10^{4} \mathrm{yr}$
C $1.867 \times 10^{9} \mathrm{yr}$
D $2 \times 10^{5} \mathrm{yr}$

The correct answer is:
$1.867 \times 10^{9} \mathrm{yr}$

Solution:
Let the initial mass of uranium be $\mathrm{M}_{0}$ Final mass of uranium after time $\mathrm{t}$

$$

\mathrm{M}=\frac{3}{4} \mathrm{M}_{0}

$$

According to the law of radioactive disintegration.

$$

\begin{array}{l}

\frac{\mathrm{M}}{\mathrm{M}_{0}}=\left(\frac{1}{2}\right)^{\mathrm{t} / \mathrm{T}} \Rightarrow \frac{\mathrm{M}_{0}}{\mathrm{M}}=(2)^{\mathrm{t} / \mathrm{T}} \\

\therefore \log _{10}\left(\frac{\mathrm{M}_{0}}{\mathrm{M}}\right)=\frac{\mathrm{t}}{\mathrm{T}} \log _{10}(2) \\

\mathrm{t}=\mathrm{T} \frac{\log _{10}\left(\frac{\mathrm{M}_{0}}{\mathrm{M}}\right)}{\log _{10}(2)}=\frac{\mathrm{T} \log _{10}\left(\frac{4}{3}\right)}{\log _{10}(2)} \\

=\frac{\mathrm{T} \log _{10}(1.333)}{\log _{10}(2)}=4.5 \times 10^{9}\left(\frac{0.1249}{0.3010}\right) \\

\Rightarrow \mathrm{t}=1.867 \times 10^{9} \mathrm{yr} .

\end{array}

$$

Topic: Physics, Nuclear Physics, BITSAT, BITSAT 2015