datavorous/entrance-exam-dataset huggingface dataset:
$A B C$ is a triangle in which $B C=10 \mathrm{~cm}, C A=6 \mathrm{~cm}$ and $A B=8 \mathrm{~cm} .$ Which one of the following is correct?

Options:
A $A B C$ is an acute angled triangle
B $A B C$ is an obtuse angled triangle
C $A B C$ is a right angled triangle
D None of these

The correct answer is:
$A B C$ is a right angled triangle

Solution:
Given a $\triangle \mathrm{ABC}$ in which $\mathrm{BC}=10 \mathrm{~cm}$, $\mathrm{CA}=6 \mathrm{~cm}$ and $\mathrm{AB}=8 \mathrm{~cm} .$
Since, $\mathrm{CA}^{2}+\mathrm{AB}^{2}=36+64=100=\mathrm{BC}^{2}$
$\therefore \Delta \mathrm{ABC}$ is a right angled triangle.

Topic: Mathematics, Properties of Triangles, NDA, NDA 2010 (Phase 2)
$A B C$ is a triangle right-angled at $B$. The hypotenuse $(A C)$ is four times the perpendicular $(B D)$ drawn to it from the opposite vertex and $A D < D C$.
What is one of the acute angle of the triangle?

Options:
A $15^{\circ}$
B $30^{\circ}$
C $45^{\circ}$
D None of these

The correct answer is:
$15^{\circ}$

Solution:
Let $\mathrm{BD}=\mathrm{P}$ and $\mathrm{DE}=x$
$\Rightarrow
\quad A C=4 P$
Let $\mathrm{E}$ be mid-point of $\mathrm{AC}$. Then, $\mathrm{AE}=\mathrm{EC}=\mathrm{BE}=2 \mathrm{P}$
In $\Delta \mathrm{BDE},(\mathrm{BE})^{2}=(\mathrm{BD})^{2}+(\mathrm{ED})^{2}$
$\Rightarrow(2 P)^{2}=(P)^{2}+x^{2}$
$\Rightarrow \quad 4 \mathrm{P}^{2}=\mathrm{P}^{2}+\mathrm{x}^{2}$
$\Rightarrow \quad 3 \mathrm{P}^{2}=\mathrm{x}^{2} \Rightarrow x=\sqrt{3} \mathrm{P}$
Now, $\mathrm{AD}=2 \mathrm{P}-x=2 \mathrm{P}-\sqrt{3} \mathrm{P}=\mathrm{P}(2-\sqrt{3})$
$\mathrm{DC}=2 \mathrm{P}+x=2 \mathrm{P}+\sqrt{3} \mathrm{P}=\mathrm{P}(2+\sqrt{3})$
In $\Delta \mathrm{BAD}, \tan \mathrm{A}=\frac{\mathrm{BD}}{\mathrm{AD}}=\frac{\mathrm{P}}{\mathrm{P}(2-\sqrt{3})}$
$=\frac{1}{2-\sqrt{3}} \times \frac{2+\sqrt{3}}{2-\sqrt{3}}=2+\sqrt{3}=\tan 75^{\circ}$
$\tan \alpha=\frac{\mathrm{AD}}{\mathrm{BD}}=\frac{\mathrm{P}(2-\sqrt{3})}{\mathrm{P}}=2-\sqrt{3}=\tan 15^{\circ}$
$\Rightarrow \quad \alpha=15^{\circ}$
As, $\Delta \mathrm{ABC}$ is right angled at $\mathrm{B}$, from figure $\alpha+\beta=90^{\circ}$ $\Rightarrow \quad 15+\beta=90^{\circ} \Rightarrow \beta=75^{\circ}$
In $\triangle \mathrm{ABC}, \angle \mathrm{A}+\angle \mathrm{B}+\angle \mathrm{C}=180^{\circ}$
$\Rightarrow \quad 75^{\circ}+90^{\circ}+\angle \mathrm{C}=180^{\circ}$
$\Rightarrow \quad \angle \mathrm{C}=180^{\circ}-165^{\circ}=15^{\circ}$
One of the a cute angle is $15^{\circ}$

Topic: Mathematics, Properties of Triangles, NDA, NDA 2011 (Phase 1)
$A B C$ is a triangle right-angled at $B$. The hypotenuse $(A C)$ is four times the perpendicular $(B D)$ drawn to it from the opposite vertex and $\mathrm{AD} < \mathrm{DC}$.
What is $\angle A B D_{?}$

Options:
A $15^{\circ}$
B $30^{\circ}$
C $45^{\circ}$
D None of these

The correct answer is:
$15^{\circ}$

Solution:
$\angle \mathrm{ABD}=\alpha=15^{\circ}$

Topic: Mathematics, Properties of Triangles, NDA, NDA 2011 (Phase 1)
$A B C$ is a triangle right-angled at $B$. The hypotenuse $(A C)$ is four times the perpendicular $(B D)$ drawn to it from the opposite vertex and $\mathrm{AD} < \mathrm{DC}$.
What is $A D: D C$ equal to?

Options:
A $(7-2 \sqrt{3}): 1$
B $(7-4 \sqrt{3}): 1$
C $1: 2$
D None of these

The correct answer is:
$(7-4 \sqrt{3}): 1$

Solution:
$\mathrm{AD}: \mathrm{DC}=\frac{\mathrm{AD}}{\mathrm{DC}}=\frac{\mathrm{P}(2-\sqrt{3})}{\mathrm{P}(2+\sqrt{3})}=\frac{2-\sqrt{3}}{2+\sqrt{3}} \times \frac{2-\sqrt{3}}{2-\sqrt{3}}$
$=\frac{4+3-4 \sqrt{3}}{1}=\frac{7-4 \sqrt{3}}{1}$
$\mathrm{AD}: \mathrm{DC}=7-4 \sqrt{3}: 1$

Topic: Mathematics, Properties of Triangles, NDA, NDA 2011 (Phase 1)
$A B C$ is a triangle right-angled at $B$. The hypotenuse $(A C)$ is four times the perpendicular $(B D)$ drawn to it from the opposite vertex and $\mathrm{AD} < \mathrm{DC}$.
What is tan $(A-C)$ equal to?

Options:
A 0
B 1
C 2
D None of these

The correct answer is:
None of these

Solution:
$\tan (A-C)=\tan \left(75^{\circ}-15^{\circ}\right)=\tan 60^{\circ}=\sqrt{3}$

Topic: Mathematics, Properties of Triangles, NDA, NDA 2011 (Phase 1)
$A B C$ is an equilateral triangle inscribed in a circle of centre $O$ and radius $5 \mathrm{~cm}$. Let the diameter through $C$ meet the circle again at $\mathrm{D}$.
Assertion (A): $A D \cdot B D < O B \cdot O C$
Reason (R): $2\left(A D^{2}+B D^{2}\right)=C D^{2}=100$ sq cm

Options:
A Both A and R are individually true and $\mathrm{R}$ is the correct explanation of $\mathrm{A}$
B Both $\mathrm{A}$ and $\mathrm{R}$ are individually true but $\mathrm{R}$ is not the correct explanation of $\mathrm{A}$
C $\mathrm{A}$ is true but $\mathrm{R}$ is false
D A is false but $\mathrm{R}$ is true

The correct answer is:
A is false but $\mathrm{R}$ is true

Solution:
(A) : Consider $A D \cdot B D < O B \cdot O C$
Now, $\mathrm{OA}=5 \mathrm{~cm}$ (Radius), $\mathrm{OB}=5 \mathrm{~cm}$ (Radius) and $\mathrm{OC}=5 \mathrm{~cm}($ Radius $)$
Since, $\Delta O A D$ and $\triangle O B D$
are congruent by SAS therefore
$$
A D=O A=5 \mathrm{~cm}
$$
and $B D=O B=5 \mathrm{~cm}$
Thus, $A D \cdot B D=5 \times 5=25$
and $\quad O B \cdot O C=5 \times 5=25$
Thus, we have
$A D \cdot B D=25=O B \cdot O C$
$$
\begin{aligned}
\operatorname{Now}(\mathrm{R}): & 2\left(A D^{2}+B D^{2}\right) \\
&=2[25+25]=100 \\
\text { and } \quad & C D^{2}=(10)^{2}=100
\end{aligned}
$$
Thus, $2\left(A D^{2}+B D^{2}\right)=C D^{2}=100$ sq. $\mathrm{cm}$.
Hence, (A) is false and (R) is true.

Topic: Mathematics, Circle, NDA, NDA 2008 (Phase 2)
A body is suspended from a string of length 1 m and mass 2 g. The mass of the body to produce a fundamental mode of 100 Hz frequency in the string is
(Acceleration due to gravity =10 m s-2)

Options:
A 80 g
B 4 kg
C 400 g
D 8 kg

The correct answer is:
8 kg

Solution:
Mass per unit length of the given string is μ=ml=2×10-31=2×10-3 kg m-1

Speed of transverse waves on a string is given by,

v=Tμ=mgμ

Fundamental frequency on a string fixed at both ends is given by,

f0=v2l=mgμ2l⇒m=4l2f02μg=4×1×104×2×10-310=8 kg

Topic: Physics, Waves and Sound, AP EAMCET, AP EAMCET 2022 (04 Jul Shift 1)
A body of density d1 is counterpoised by Mg of weights of density d2 in the air of density d. Then the true mass of the body is

Options:
A M

B M=1-dd2
C M=1-dd1
D M1-d/d21-d/d1

The correct answer is:
M1-d/d21-d/d1

Solution:
Let M0 = mass of body in vacuum.
Apparent weight of the body in air=Apparent weight of standard weights in air
â Actual weight - upthrust due to displaced air = Actual weight - upthrust due to displaced air
â M 0 g - M 0 d 1 d g = M g - M d 2 dg â M 0 = M 1 - d d 2 1 - d d 1

Topic: Physics, Mechanical Properties of Fluids, NEET
A body of density d, is counterpoised by Mg of weight of density d2 in air of density d . Then, the true mass of the body is

Options:
A M
B M1-dd2
C M1-dd1
D M1-dd21-dd1

The correct answer is:
M1-dd21-dd1

Solution:
Let the M1 is the mass of the body. Apparent weight of the body in air
= Apparent weight of standard weight in air
∴ Actual weight - Upthrust due to displaced air
⇒M1g-M1d1dg=Mg-Md2dg
M1g1-dd1=Mg1-dd2
∴M1=M1-dd21-dd1

Topic: Physics, Mechanical Properties of Fluids, NEET
A body of mass $0.3 \mathrm{~kg}$ hangs by a spring with a force constant of $50 \mathrm{~N} / \mathrm{m}$. The amplitude of oscillations is damped and reaches $\frac{1}{e}$ of its original value in about 100 oscillations. If $\omega$ and $\omega^{\prime}$ are the angular frequencies of undamped and damped oscillations respectively, then percentage of $\left(\frac{\omega-\omega}{\omega}\right)$ is

Options:
A $\left(\frac{1}{800 \pi}\right)$
B $\left(\frac{\pi^2}{600}\right)$
C $\left(\frac{1}{800 \pi^2}\right)$
D $\left(\frac{\pi}{400}\right)$

The correct answer is:
$\left(\frac{1}{800 \pi^2}\right)$

Solution:
Hint The displacement of a spring mass oscillator, $X=A e^{-b t / 2 m} \cos \left(\omega^{\prime} t+\phi\right)$ where, $\omega^{\prime}=$ angular frequency of damped oscillation and $\omega^{\prime}=\omega_0 \sqrt{1-\frac{b^2}{4 m^2 \omega_0^2}}$

Topic: Physics, Oscillations, TS EAMCET, TS EAMCET 2019 (04 May Shift 1)
A body of mass M hits normally a rigid wall with velocity V and bounces back with the same velocity. The impulse experienced by the body is :

Options:
A

MV

B

1.5 MV

C

2 MV

D Zero

The correct answer is:

2 MV

Solution:
Impulse, J → = ∫ t 1 t 2 F → .dt

or J= F avg . Δt

Now, according to Newton’s Second Law

F → = d P → dt   or   ∫ t 1 t 2 F → . dt = ∫ P → 1 P → 2 d P →

∴   J → = P → 2 − P → 1 =Δ P →

[i.e., the action of impulse is to change the momentum of a body and the impulse of force is equal to the change in momentum. This is also known as impulse-momentum theorem]

In the given case,

Impulse = p → 2 − p → 1
(i.e., impulse = change in momentum)

or Impulse, J=M  v → 2 −M  v → 1

⇒  J=M[(v)−(−v)]⇒  J=2Mv   N s

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

Topic: Physics, Center of Mass Momentum and Collision, NEET
A body of mass $m$ is situated in a potential field $U(x)=U_0(1-\cos \alpha x)$ when, $U_0$ and $\alpha$ are constants. Find the time period of small oscillations.

Options:

None

Solution:
As given that, the potential energy (associated with the field) for a body of mass $m$.
$$
U(x)=U_0(1-\cos \alpha x)
$$
$\because$ for conservative force $f$, we can write $f=\frac{-d u}{d x}$
We have assumed the field to be conservative
$$
\begin{aligned}
F &=-\frac{d}{d x}\left(U_0-U_0 \cos \alpha x\right) \\
&=-U_0 \alpha \sin \alpha x
\end{aligned}
$$
For SHM, $\alpha x$ is small so $\sin \alpha x$ i.e. $\sin \alpha x=\alpha x$ so,
$$
\begin{aligned}
&F=-U_0 \alpha^2 x \\
&\therefore F \propto(-x)
\end{aligned}
$$
As, $U_0, \alpha$ being constant.
$\therefore$ So motion is SHM for small oscillations.
Standard equation for SHM,
$$
F=-m \omega^2 x
$$
Comparing Eqs. (ii) and (iii), then
$$
\begin{aligned}
&-m \omega^2 x=-U_0 \alpha^2 x \\
&m \omega^2=U_0 \alpha^2 \\
&\omega^2=\frac{U_0 \alpha^2}{m} \text { or } \omega=\sqrt{\frac{U_0 \alpha^2}{m}} \\
&\therefore \text { Time period } T=\frac{2 \pi}{\omega}=2 \pi \sqrt{\frac{m}{U_0 \alpha^2}}
\end{aligned}
$$
From (i) this period is valid for small angle $\alpha x$.
$$
T=\frac{2 \pi}{\alpha} \sqrt{\frac{m}{U_0}}
$$

Topic: Physics, Oscillations
A box contains 3 white and 2 black balls. Two balls are drawn at random one after the other. If the balls are not replaced, what is the probability that both the balls are black?

Options:
A $2 / 5$
B $1 / 5$
C $1 / 10$
D None of these

The correct answer is:
$1 / 5$

Solution:
Total number of balls $=5$ Number of black balls $=2$ Required probability $=\frac{\mathrm{n}(\mathrm{E})}{\mathrm{n}(\mathrm{S})}=\frac{{ }^{2} \mathrm{C}_{1} \times{ }^{1} \mathrm{C}_{1}}{{ }^{5} \mathrm{C}_{2}}=\frac{2}{5} \times \frac{1}{2}=\frac{1}{5}$

Topic: Mathematics, Probability, NDA, NDA 2014 (Phase 1)
A box contains 6 distinct dolls. From this box, 3 dolls are randomly selected one by one with replacement. What is the probability of selecting 3 distinct dolls?

Options:
A $5 / 54$
B $12 / 25$
C $1 / 20$
D $5 / 9$

The correct answer is:
$5 / 9$

Solution:
Required probability $=\frac{{ }^{6} \mathrm{C}_{1} \times{ }^{5} \mathrm{C}_{1} \times{ }^{4} \mathrm{C}_{1}}{{ }^{6} \mathrm{C}_{1} \times{ }^{6} \mathrm{C}_{1} \times{ }^{6} \mathrm{C}_{1}}$
$=\frac{6 \times 5 \times 4}{6 \times 6 \times 6}=\frac{5}{9}$

Topic: Mathematics, Probability, NDA, NDA 2011 (Phase 1)
A box contains 6 pens, 2 of which are defective. Two pens are taken randomly from the box. If random variable x: Number of defective pens obtained, then standard deviation of x=

Options:
A ± 43√5
B 83
C 1645
D 435

The correct answer is:
435

Solution:
x : no. of defective pens
Two pens are taken from box
â´ x can taken values 0, 1, 2
px=0= 4C2 6C2=4Ã36Ã5=25=615
p x=1= 2C1Ã 4C1 6C2=2Ã4Ã2Ã16Ã5=815
px=2= 2C2 6C2=1Ã2Ã16Ã5=115

x	P	xipi	xi2pi

0	615	0	0

1	815	815	815

2	115	215	415

Ex=1015
=23
Ex2=1215
=45
Standard deviation =Ex2-Ex2
Standard deviation =45-232
=45-49
=4Ã445
=43â5

Topic: Mathematics, Probability, MHT CET, MHT CET 2017
A box contains 10 identical electronic components of which 4 are defective. If 3 components are selected at random from the box in succession, without replacing the units already drawn, what is the probability that two of the selected components are defective?

Options:
A $1 / 5$
B $5 / 24$
C $3 / 10$
D $1 / 40$

The correct answer is:
$3 / 10$

Solution:
Total number of selecting 3 components out of $10=$ ${ }^{10} \mathrm{C}_{3}$. Out of 3 selected components two defective pieces can be selected in ${ }^{4} \mathrm{C}_{2}$ ways and one non-defective piece will be selected in ${ }^{6} \mathrm{C}_{1}$ ways, hence.
Required probability $=\frac{{ }^{6} \mathrm{C}_{1} \times{ }^{4} \mathrm{C}_{2}}{{ }^{10} \mathrm{C}_{3}}=\frac{6 \times 6 \times 6}{10 \times 9 \times 8}=\frac{3}{10}$

Topic: Mathematics, Probability, NDA, NDA 2007 (Phase 1)
A boy throws a cricket ball from the boundary to the wicket keeper. If the frictional force due to air fa cannot be ignored, the forces acting on the ball at the position X are represented by

Options:
A
B
C
D

The correct answer is:

Solution:
The fiction force due to air on the ball fa will always act in the opposite direction of the velocity of the ball, and the velocity of the ball at any point on the path is always in the tangential direction of the path of the projectile.

The weight of the body always acts downward perpendicular to the surface of the earth and along the centre of the earth.

Topic: Physics, Laws of Motion, AP EAMCET, AP EAMCET 2021 (20 Aug Shift 1)
A boy throws a cricket ball from the boundary to the wicket-keeper. If the frictional force due to air cannot be ignored, the forces acting on the ball at the position $X$ are respected by

Options:
A

B

C

D

The correct answer is:

Solution:
The forces acting on the ball will be (i) in the direction opposite to its motion ie, frictional force and (ii) weight mg.

Topic: Physics, Laws of Motion, KCET, KCET 2010
A can hit a target 4 times in 5 shots; $\mathrm{B}$ can hit a target 3 times in 4 shots:
$\mathrm{C}$ can hit a target 2 times in 3 shots All the three fire a shot each. What is the probability that two shots are at least hit?

Options:
A $1 / 6$
B $3 / 5$
C $5 / 6$
D $1 / 3$

The correct answer is:
$5 / 6$

Solution:
Probability of no one hitting the target
$=\frac{1}{5 \times 4 \times 3}=\frac{1}{60}$
Probability of one hitting the target
$=\frac{4+3+2}{60}=\frac{9}{60}$
$\therefore$ Probability of maximum one hit
$=\frac{1}{60}+\frac{9}{60}=\frac{10}{60}=\frac{1}{6}$
Probability that two shots are hit at least is the required
probability $=1-\frac{1}{6}=\frac{5}{6}$

Topic: Mathematics, Probability, NDA, NDA 2007 (Phase 1)
A car manufacturing factory has two plants $X$ and $Y$. Plant $X$ manufactures $70 \%$ of cars and plant $Y$ manufactures $30 \%$ of cars. $80 \%$ of cars at plant $X$ and $90 \%$ of cars at plant $Y$ are rated as standard quality. A car is chosen at random and is found to be standard quality. The probability that it has come from plant $X$ is

Options:
A $\frac{56}{73}$
B $\frac{56}{84}$
C $\frac{56}{83}$
D $\frac{56}{79}$

The correct answer is:
$\frac{56}{83}$

Solution:
Let $E=$ the event that the car is of standard quảlity.
Let $A_{1}=$ the event that the car is manufactured in plant $X$.
and $A_{2}=$ the event that the car is manufactured in plant $Y$.
Now, $P\left(A_{1}\right)=\frac{70}{100}=\frac{7}{10}, P\left(A_{2}\right)=\frac{30}{100}=\frac{3}{10}$ $P\left(\frac{E}{A_{1}}\right)=$ Probability that a standard quality car is manufactured in plant $X=\frac{80}{100}=\frac{8}{10}$ $P\left(\frac{E}{A_{2}}\right)=$ Probability that a standard quality car is manu factured in blant $Y=\frac{90}{100}=\frac{9}{10}$ $P\left(\frac{A_{1}}{E}\right)=$ Probability that a standard quality car has come from plant $X$
$\begin{aligned}
=& \frac{P\left(A_{1}\right) \times P\left(\frac{E}{A_{1}}\right)}{P\left(A_{1}\right) \cdot P\left(\frac{E}{A_{1}}\right)+P\left(A_{2}\right) \cdot P\left(\frac{E}{A_{2}}\right)} \\
=& \frac{\frac{7}{10} \times \frac{8}{10}}{\frac{7}{10} \times \frac{8}{10}+\frac{3}{10} \times \frac{9}{10}} \\
=& \frac{7 \times 8}{7 \times 8+3 \times 9} \\
=& \frac{56}{56+27}=\frac{56}{83}
\end{aligned}$
Hence, the required probability is $\frac{56}{83}$.

Topic: Mathematics, Probability, KCET, KCET 2021
A card is drawn from a pack of 52 cards and a gambler bets that it is a spade or an ace. Which one of the following are the odds against his winning this bet?

Options:
A 13 to 4
B 4 to 13
C 9 to 4
D 4 to 9

The correct answer is:
9 to 4

Solution:
Probability of getting a spade $=\frac{13}{52}$
Probability of an ace $=\frac{4}{52}$
and probability of getting a spade ace $=\frac{1}{52}$
$\therefore \quad$ Required probability $=\frac{13}{52}+\frac{4}{52}-\frac{1}{52}=\frac{16}{52}=\frac{4}{13}$
Odds against his winning $=\frac{1-\frac{4}{13}}{\frac{4}{13}}=\frac{\frac{9}{13}}{\frac{4}{13}}=\frac{9}{4}$

Topic: Mathematics, Probability, NDA, NDA 2007 (Phase 1)
A card is drawn from a well-shuffled ordinary deck of 52 cards. What is the probability that it is an ace?

Options:
A $\frac{1}{13}$
B $\frac{2}{13}$
C $\frac{3}{13}$
D $\frac{1}{52}$

The correct answer is:
$\frac{1}{13}$

Solution:
$\quad$ Number of ace cards $=4$
So, probability $=\frac{4}{52}=\frac{1}{13}$

Topic: Mathematics, Probability, NDA, NDA 2017 (Phase 1)
A card is drawn from a wel-shuffled deck of 52 cards. What is the probability that it is queen of spade?

Options:
A $\frac{1}{52}$
B $\frac{1}{13}$
C $\frac{1}{4}$
D $\frac{1}{8}$

The correct answer is:
$\frac{1}{52}$

Solution:
Prob. of getting queen of spade $=\frac{{ }^{1} C_{1}}{{ }^{52} C_{1}}=\frac{1}{52}$

Topic: Mathematics, Probability, NDA, NDA 2016 (Phase 1)
A Cassegrain telescope uses two mirrors as shown in figure. Such a telescope is built with the mirrors 20 $\mathrm{mm}$ apart. If the radius of curvature of the large mirror is $220 \mathrm{~mm}$ and the small mirror is $140 \mathrm{~mm}$, where will the final image of an object at infinity be?

Options:

None

Solution:
For objective mirror (concave), $u=-\infty$
Radius of curvature $R=-220 \mathrm{~mm}=-22 \mathrm{~cm}$
Focal length $f=\frac{R}{2}=-11 \mathrm{~cm}$
Image will be obtained at focus.
$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$ or $\frac{1}{v}+\frac{1}{-\infty}=\frac{1}{-11}$ or $v=-11 \mathrm{~cm}$
Image formed by objective mirror act as a virtual object
for secondary mirror.
Distance of virtual object for secondary mirror.
$u_0=(11-2) \mathrm{cm}=+9 \mathrm{~cm}$
Focal length of convex mirror used $=14 \mathrm{~cm}$
So, $\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$ or $\frac{1}{v}=\frac{1}{14}-\frac{1}{9}$ or $v=-25.5 \mathrm{~cm}$

Topic: Physics, Ray Optics
A cell in secondary circuit gives null deflection for $2.5 \mathrm{~m}$ length of potentiometer having $10 \mathrm{~m}$ length of wire. If the length of the potentiometer wire is increased by $1 \mathrm{~m}$ without changing the cell in the primary, the position of the null point now is

Options:
A 3.5 m
B 3 m
C 2.75 m
D 2.0 m

The correct answer is:
3.5 m

Solution:
Resistance of potentiometer wire
$R=\rho \times \frac{l}{A}$
or $\quad R=\left(\rho \times \frac{10}{A}\right)$
The value of $2.5 \mathrm{~m}$ length wire
$R^{\prime}=\frac{\rho \times 10}{A \times 10} \times 2.5$
or $\quad R^{\prime}=\left(\frac{2.5 \rho}{A \times 10}\right)$
Potential
$\begin{aligned}
V^{\prime} & =I \times R^{\prime} \\
& =I\left(\frac{2.5 \rho}{A \times 10}\right)
\end{aligned}$
Now, again the length of potentiometer wire is increased by $1 \mathrm{~m}$, then resistance of null position wire.
$\begin{aligned}
R^{\prime \prime} & =\left(\frac{\rho \times l}{11 \times A}\right) \\
V^{\prime \prime} & =I R^{\prime \prime} \\
and \quad V & =V^{\prime}
\end{aligned}$
$\begin{aligned} \frac{I \times 2.5 \rho}{A \times 10} & =\frac{\rho \times l}{11 \times A} \times I \\ \text { or } \quad \frac{2.5 \times 11}{10} & =l=2.75 \mathrm{~m}\end{aligned}$

Topic: Physics, Current Electricity, TS EAMCET, TS EAMCET 2009
A cell in secondary circuit gives null deflection for $2.5 \mathrm{~m}$ length of potentiometer having $10 \mathrm{~m}$ length of wire. If the length of the potentiometer wire is increased by $1 \mathrm{~m}$ without changing the cell in the primary, the position of the null point now is

Options:
A 3.5 m
B 3 m
C 2.75 m
D 2.0 m

The correct answer is:
3.5 m

Solution:
Resistance of potentiometer wire
$R=\rho \times \frac{l}{A}$
or $\quad R=\left(\rho \times \frac{10}{A}\right)$
The value of $2.5 \mathrm{~m}$ length wire
$R^{\prime}=\frac{\rho \times 10}{A \times 10} \times 2.5$
or $\quad R^{\prime}=\left(\frac{2.5 \rho}{A \times 10}\right)$
Potential
$\begin{aligned}
V^{\prime} & =I \times R^{\prime} \\
& =I\left(\frac{2.5 \rho}{A \times 10}\right)
\end{aligned}$
Now, again the length of potentiometer wire is increased by $1 \mathrm{~m}$, then resistance of null position wire.
$\begin{aligned}
R^{\prime \prime} & =\left(\frac{\rho \times l}{11 \times A}\right) \\
V^{\prime \prime} & =I R^{\prime \prime} \\
and \quad V & =V^{\prime}
\end{aligned}$
$\begin{aligned} \frac{I \times 2.5 \rho}{A \times 10} & =\frac{\rho \times l}{11 \times A} \times I \\ \text { or } \quad \frac{2.5 \times 11}{10} & =l=2.75 \mathrm{~m}\end{aligned}$

Topic: Physics, Current Electricity, AP EAMCET, AP EAMCET 2009
A cell in secondary circuit gives null deflection for $2.5 \mathrm{~m}$ length of potentiometer having $10 \mathrm{~m}$ length of wire. If the length of the potentiometer wire is increased by $1 \mathrm{~m}$ without changing the cell in the primary, the position of the null point now is

Options:
A $3.5 \mathrm{~m}$
B $3 \mathrm{~m}$
C $2.75 \mathrm{~m}$
D $2.0 \mathrm{~m}$

The correct answer is:
$2.75 \mathrm{~m}$

Solution:
Here length of potentiometer wire, $l=10 \mathrm{~m}$ Resistance of potentiometer wire
$R=\rho \times \frac{l}{A}$ or $R=\left(\rho \times \frac{10}{A}\right)$
The value of $2.5 \mathrm{~m}$ length wire
$$
R^{\prime}=\frac{\rho \times 10}{A \times 10} \times 2.5 \Rightarrow R^{\prime}=\left(\frac{2.5 \rho}{A \times 10}\right)
$$
Potential, $V^{\prime}=I \times R^{\prime}=I\left(\frac{2.5 \rho}{A \times 10}\right)$
Again the length of potentiometer wire is increased by $1 \mathrm{~m}$.
Resistance of null position
$$
R^{n}=\left(\frac{\rho \times l}{11 \times A}\right)
$$
$\therefore V^{\prime \prime}=I R^{\prime \prime}$ and $V=V^{\prime}$
$$
\begin{array}{l}
\Rightarrow \frac{I \times 2.5 \rho}{A \times 10}=\frac{\rho \times I}{11 \times A} \times I \\
\text { or } \frac{2.5 \times 11}{10}=I=2.75 \mathrm{~m}
\end{array}
$$

Topic: Physics, Current Electricity, VITEEE, VITEEE 2009
A certain type of missile hits the target with probability $\mathrm{p}=0.3$. What is the least number of missiles should be fired so that there is at least an $80 \%$ probability that the target is hit?

Options:
A 5
B 6
C 7
D None of the above

The correct answer is:
5

Solution:
Probability of hitting the target $=0.3$ If'n' is the no. of times that the Missile is fired. $\therefore$ Probability ofhitting at least once $=1-[1-0.3]^{\mathrm{n}}=0.8$ $0.7^{\mathrm{n}}=0.2$
$n \log 0.7=\log 0.2$
$\Rightarrow \mathrm{n}=4.512$
for $\mathrm{n}=4 ; \mathrm{p} < 0.8$
take $\mathrm{n}=5$
$n=5$
Hence 5 missiles should be fired so that there is at least $80 \%$ prob. that the target is hit.

Topic: Mathematics, Probability, NDA, NDA 2016 (Phase 1)