datavorous/entrance-exam-dataset huggingface dataset:
A student is asked to answer 10 out of 13 questions is an examination such that he must answer atleast four questions from the first five questions. The number of choices available to him is

Options:
A 140
B 176
C 196
D 280

The correct answer is:
196

Solution:
Case I
4 questions from first 5 questions and 6 questions from remaining 8 questions
$$
\therefore \quad{ }^5 C_4 \times{ }^8 C_6=5 \times 28=140
$$
Case II
Choose 5 questions from first 5 questions and next 5 questions from remaining 8 questions
$$
\because \quad{ }^5 C_5 \times{ }^8 C_5=1 \times 56=56
$$
Total number $=140+56=196$

Topic: Mathematics, Permutation Combination, TS EAMCET, TS EAMCET 2019 (04 May Shift 1)
A student is performing the experiment of the resonance column. The diameter of the column tube is 6 cm. The frequency of the tuning fork is 504 Hz. Speed of the sound at the given temperature is 336 m s-1. The zero of the meter scale coincides with the top end of the resonance column tube. The reading of the water level in the column when the first resonance occurs is:

Options:
A 14.8 cm
B 18.4 cm
C 16.6 cm
D 13 cm

The correct answer is:
14.8 cm

Solution:
d=6 cm, f=504, v=336 m s-1

e=0.3 d

l+e=λ4=V4f

l=16.66-0.3×6

l=14.866 cm

l=14.8 cm

Topic: Physics, Waves and Sound, JEE Main, JEE Main 2021 (25 Feb Shift 1)
A student is performing the resonance column experiment using a tube of diameter 4 cm and a tuning fork of frequency 512 Hz. The air temperature is 38 °C in which the speed of sound is 336 m s-1. The zero of the meter scale coincides with the top end of the resonance column tube. When the first resonance occurs, the reading of the water level in the column is

Options:
A 14.0 cm
B 15.2 cm
C 16.4 cm
D 17.6 cm

The correct answer is:
15.2 cm

Solution:
Let L=Length of air column
and e=end correction
L + e = Î» 4
using v=fÎ»
Î»=336512
=0.656 m
=65.6 cm
âL=Î»4-e e=0.3Ãdiameter
â L = Î» 4 - e
=16.4-1.2=15.2 cm

Topic: Physics, Waves and Sound, JEE Main
A student is to answer 10 out of 13 questions in an examination such that he must choose at least 4 from the first five questions. The number of choices available to him is

Options:
A $196$
B $140$
C $168$
D $176$

The correct answer is:
$196$

Solution:
Total number of questions $=13$
Now of questions to be selected $=10$
Restriction - At least 4 from first 5 questions are to be selected.
Case I If exactly 4 out of first 5 questions are selected.
Number of ways of selecting 4 questions from 5 questions $={ }^5 C_4=5$
Remaining $10-4=6$ questions are selected from last $13-5=8$ questions in ${ }^8 C_6$ ways -
$\therefore$ Number of ways $=5 \times{ }^8 C_6=140$
Case II If exactly 5 out of first 5 questions are selected.
Number of ways of selecting 5 out of 5 questions $={ }^5 C_5=1$
Remaining $10-5=5$ questions are selected from last $13-5=8$ questions in ${ }^8 C_5$ ways.
$\therefore$ Number of ways $=1 \times{ }^8 C_5=56$
$\therefore$ Total number of ways $=140+56=196$

Topic: Mathematics, Permutation Combination, AP EAMCET, AP EAMCET 2022 (08 Jul Shift 2)
A student is to answer 10 out of 13 questions in an examination such that he must choose at least 4 from the first five questions. The number of choices available to him is

Options:
A
346

B
140

C
196

D
280

The correct answer is:
196

Solution:
As for given question two cases are possible.
(i) Selecting 4 out of first five question and 6 out of remaining 8 question $={ }^5 \mathrm{C}_4 \times{ }^8 \mathrm{C}_6=140$ choices.
(ii) Selecting 5 out of first five question and 5 out of remaining 8 questions $={ }^5 \mathrm{C}_5 \times{ }^8 \mathrm{C}_5=56$ choices.
Therefore, total number of choices $=140+56=196$.

Topic: Mathematics, Permutation Combination, JEE Main, JEE Main 2003
A student measured the diameter of a small steel ball using a screw gauge of least count 0.001 cm. The main scale reading is 5 mm and 25th division of circular scale coincides with reference line. If screw gauge has a zero error of -0.004 cm, the correct diameter of the ball is

Options:
A 0.053 cm
B 0.525 cm
C 0.521 cm
D 0.529 cm

The correct answer is:
0.529 cm

Solution:
Diameter of the steel ball = main scale reading + (circular scale reading × Least count) - zero error
=MSR+( CSR×L.C )− zero error
=0.5 cm+25×0.001−( −0.004 )
=0.5+0.025+0.004
=0.529 cm

Topic: Physics, Experimental Physics, NEET, NEET 2018
A student measured the diameter of a wire using a screw gauge with the least count $0.001 \mathrm{~cm}$ and listed the measurements. The measured value should be recorded as

Options:
A
$5.3200 \mathrm{~cm}$

B
$5.3 \mathrm{~cm}$

C
$5.32 \mathrm{~cm}$

D
$5.320 \mathrm{~cm}$

The correct answer is:
$5.320 \mathrm{~cm}$

Solution:
The least count (L.C.) of a screw guage is the smallest length which can be measured accurately with it.
As least count is $0.001 \mathrm{~cm}=\frac{1}{1000} \mathrm{~cm}$
Hence measured value should be recorded upto 3 decimal places i.e., $5.320 \mathrm{~cm}$

Topic: Physics, Units and Dimensions, JEE Main, JEE Main 2012 (12 May Online)
A student measured the length of a rod and wrote it as $3.50 \mathrm{~cm}$. Which instrument did he use to measure it?

Options:
A A screw gauge having 100 divisions in the circular scale and pitch as $1 \mathrm{~mm}$
B A screw gauge having 50 divisions in the circular scale and pitch as $1 \mathrm{~mm}$
C A meter scale
D A vernier calliper where the 10 divisions in vernier scale match with 9 division in the main scale and main scale have 10 divisions in $1 \mathrm{~cm}$

The correct answer is:
A vernier calliper where the 10 divisions in vernier scale match with 9 division in the main scale and main scale have 10 divisions in $1 \mathrm{~cm}$

Solution:
The measured distance by the student is $3.50 \mathrm{~cm}$ which means the least count of the measurement done is $0.01 \mathrm{~cm}$.
The least count of a vernier calliper which has 10 divisions in lcm on the main scale and 10 divisions in vernier scale match with 9 division in the main scale can be given as
$1 \mathrm{~cm} /(10 \times 10)=0.01 \mathrm{~cm}$
Therefore, the student must have used the vernier calliper for the measurement
Answer: (4) A vernier calliper where the 10 divisions in vernier scale match with 9 divisions in the main scale and main scale have 10 divisions in $1 \mathrm{~cm}$

Topic: Physics, Mathematics in Physics, JEE Main
A student measured the length of a rod and wrote it as 3.50 cm. Which instrument did he use to measure it ?

Options:
A A meter scale
B A vernier calliper where the 10 divisions in vernier scale matches with 9 division in main scale and main scale has 10 divisions in 1 cm
C A screw gauge having 100 divisions in the circular scale and pitch as 1 mm
D A screw gauge having 50 divisions in the circular scale and pitch as 1 mm

The correct answer is:
A vernier calliper where the 10 divisions in vernier scale matches with 9 division in main scale and main scale has 10 divisions in 1 cm

Solution:
As per given data the Vernier has a least count of 0.01 cm, which is consistent with the information given in question.

Topic: Physics, Experimental Physics, JEE Main, JEE Main 2014 (06 Apr)
A student measures the distance covered and time taken by a body under free fall which was initially at rest. He uses this data for estimating g, the acceleration due to gravity. If the maximum percentage errors in measurement of the distance and the time are e1 and e2 respectively, then the maximum possible percentage error in the estimation of g is

Options:
A e2−e1
B e1+ 2e2
C e1+e2
D e1−2e2

The correct answer is:
e1+ 2e2

Solution:
From the relation
h = ut + 1 2 gt 2
h = 1 2 gt 2 ⇒ g = 2 h t 2 ∵ body initially at rest
Taking natural logarithms on both sides, we get
lng=lnh-2lnt+ln2
Differentiating, Δ g g = Δ h h - 2 Δ t t
For maximum permissible errors, we add both the errors. So
Δgg×100max=Δhh×100+2×Δtt×100
Δ h h × 100 = e 1 and Δ t t × 100 = e 2
Therefore, Δ g g × 100 max = e 1 + 2 e 2

Topic: Physics, Experimental Physics, JEE Main
A student measures the distance traversed in free fall of a body, initially at rest in a given time. He uses this data to estimate $g$, the acceleration due to gravity. If the maximum percentage errors in measurement of the distance and the time are $e_1$ and $e_2$ respectively, the percentage error in the estimation of $g$ is

Options:
A $\mathrm{e}_2-\mathrm{e}_1$
B $e_1+2 e_2$
C $e_1+e_2$
D $e_1-2 e_2$

The correct answer is:
$e_1+2 e_2$

Solution:
$$
\begin{aligned}
& h=\frac{1}{2} g t^2 \\
& \text { or } \quad g=\frac{\mathrm{h}}{\mathrm{t}^2} \\
& \therefore \quad \log \mathrm{g}=\log \mathrm{h}-2 \log \mathrm{t} \\
& \left(\frac{\Delta g}{g} \times 100\right)_{\max }=\left(\frac{\Delta \mathrm{h}}{\mathrm{h}} \times 100\right)+2\left(\frac{\Delta \mathrm{t}}{\mathrm{t}} \times 100\right) \\
& =\mathrm{e}_1+2 \mathrm{e}_2 \\
&
\end{aligned}
$$

Topic: Physics, Mathematics in Physics, JEE Main
A student measures the distance traversed in free fall of a body, initially at rest in a given time. He uses this data to estimate $g$, the acceleration due to gravity. If the maximum percentage errors in measurement of the distance and the time are $e_1$ and $e_2$ respectively, the percentage error in the estimation of $g$ is

Options:
A $\mathrm{e}_2-\mathrm{e}_1$
B $e_1+2 e_2$
C $e_1+e_2$
D $e_1-2 e_2$

The correct answer is:
$e_1+2 e_2$

Solution:
$$
\begin{aligned}
& h=\frac{1}{2} g t^2 \\
& \text { or } \quad g=\frac{\mathrm{h}}{\mathrm{t}^2} \\
& \therefore \quad \log \mathrm{g}=\log \mathrm{h}-2 \log \mathrm{t} \\
& \left(\frac{\Delta g}{g} \times 100\right)_{\max }=\left(\frac{\Delta \mathrm{h}}{\mathrm{h}} \times 100\right)+2\left(\frac{\Delta \mathrm{t}}{\mathrm{t}} \times 100\right) \\
& =\mathrm{e}_1+2 \mathrm{e}_2 \\
&
\end{aligned}
$$

Topic: Physics, Mathematics in Physics, NEET, NEET 2010 (Mains)
A student measures the focal length of convex lens by putting an object pin at a distance 'u' from the lens and measuring the distance ' $v$ ' of the image pin. The graph between ' $u$ ' and ' $v$ ' plotted by the student should look like

Options:
A

B

C

D

The correct answer is:

Solution:
$$
\frac{1}{v}-\frac{1}{u}=\frac{1}{f}=\text { constant }
$$

Topic: Physics, Ray Optics, JEE Main, JEE Main 2008
A student measures the terminal potential difference (V) of a cell (of emf $\varepsilon$ and internal) resistance $r$ ) as a function of the current (I) flowing through it. The slope and intercept of the graph between $\mathrm{V}$ and $\mathrm{I}$, then respectively equal to :

Options:
A $-\epsilon$ and $r$
B $\in$ and $-r$
C $-\mathrm{r}$ and $\in$
D $r$ and $-\epsilon$

The correct answer is:
$-\mathrm{r}$ and $\in$

Solution:
$$
\begin{aligned}
& \quad \mathrm{E}=\mathrm{V}+\mathrm{Ir} \\
& \Rightarrow \mathrm{V}=\mathrm{E}-\mathrm{Ir} \\
& \text { Comparing with } \mathrm{y}=\mathrm{mx}+\mathrm{c} \\
& \text { Slope }=-\mathrm{r} \text {, intercept }=\mathrm{E}
\end{aligned}
$$

Topic: Physics, Current Electricity, JEE Main
A student measures the terminal potential difference (V) of a cell (of emf $\varepsilon$ and internal) resistance $r$ ) as a function of the current (I) flowing through it. The slope and intercept of the graph between $\mathrm{V}$ and $\mathrm{I}$, then respectively equal to :

Options:
A $-\epsilon$ and $r$
B $\in$ and $-r$
C $-\mathrm{r}$ and $\in$
D $r$ and $-\epsilon$

The correct answer is:
$-\mathrm{r}$ and $\in$

Solution:
$$
\begin{aligned}
& \quad \mathrm{E}=\mathrm{V}+\mathrm{Ir} \\
& \Rightarrow \mathrm{V}=\mathrm{E}-\mathrm{Ir} \\
& \text { Comparing with } \mathrm{y}=\mathrm{mx}+\mathrm{c} \\
& \text { Slope }=-\mathrm{r} \text {, intercept }=\mathrm{E}
\end{aligned}
$$

Topic: Physics, Current Electricity, NEET, NEET 2009 (Mains)
A student measures the time period of 100 oscillations of a simple pendulum four times. The data set is 90 s, 91 s, 95 s and 92 s. If the minimum division in the measuring clock is 1 s, then the reported mean time should be:

Options:
A 92Â±1.8Â s
B 92Â±3Â s
C 92Â±2 s
D 92Â±5.0Â s

The correct answer is:
92Â±2 s

Solution:
T	T s	T i - T

t 1	90	- 2

t 2	91	- 1

t 3	95	3

t 4	92	0

T i	92	Σ|Ti-T|N=1.5

Tr = 9 2 ± 1.5
Tr = 9 2 ± 2
Because least count of clock is 1s

Topic: Physics, Mathematics in Physics, JEE Main
A student measures the time period of 100 oscillations of a simple pendulum four times. The data set is 90 s, 91 s, 95 s and 92 s. If the minimum division in the measuring clock is 1 s, then the reported mean time should be:

Options:
A 92Â±1.8Â s
B 92Â±3Â s
C 92Â±2 s
D 92Â±5.0Â s

The correct answer is:
92Â±2 s

Solution:
T	T s	T i - T

t 1	90	- 2

t 2	91	- 1

t 3	95	3

t 4	92	0

T i	92	Σ|Ti-T|N=1.5

Tr = 9 2 ± 1.5
Tr = 9 2 ± 2
Because least count of clock is 1s

Topic: Physics, Mathematics in Physics, JEE Main, JEE Main 2016 (03 Apr)
A student needs to prepare a buffer solution of propanoic acid and its sodium salt with pH4. The ratio of CH3CH2COO-CH3CH2COOH required to make buffer is Given : KaCH3CH2COOH=1.3×10-5

Options:
A 0.03
B 0.13
C 0.23
D 0.33

The correct answer is:
0.13

Solution:
Henderson's equation for calculating pH of acidic buffer solutions is given as

pH=pKa+logSaltAcid

pH=pKa+logCH3COO-CH3COOH

4=5-log(1.3)+logCH3COO-CH3COOH

log1.3×10-1=logCH3COO-CH3COOH

CH3COO-CH3COOH=0.13

Topic: Chemistry, Ionic Equilibrium, JEE Main, JEE Main 2022 (28 Jun Shift 2)
A student performs an experiment to determine the Young's modulus of a wire, exactly $2 \mathrm{~m}$ long, by Searle's method. In a particular reading, the student measures the extension in the length of the wire to be $0.8 \mathrm{~mm}$ with an uncertainty of $\pm 0.05 \mathrm{~mm}$ at a load of exactly $1.0 \mathrm{~kg}$. The student also measures the diameter of the wire to be $0.4 \mathrm{~mm}$ with an uncertainty of $\pm 0.01 \mathrm{~mm}$. Take $g=9.8 \mathrm{~m} / \mathrm{s}^2$ (exact). The Young's modulus obtained from the reading is

Options:
A
$(2.0 \pm 0.3) 10^{11} \mathrm{~N} / \mathrm{m}^2$

B
$(2.0 \pm 0.2) \times 10^{11} \mathrm{~N} / \mathrm{m}^2$

C
$(2.0 \pm 0.1) \times 10^{11} \mathrm{~N} / \mathrm{m}^2$

D
$(2.0 \pm 0.05) \times 10^{11} \mathrm{~N} / \mathrm{m}^2$

The correct answer is:
$(2.0 \pm 0.2) \times 10^{11} \mathrm{~N} / \mathrm{m}^2$

Solution:
$\begin{aligned} Y & =\frac{F L}{A l}=\frac{4 F L}{\pi d^2 l} \\ & =\frac{(4)(1.0 \times 9.8)(2)}{\pi\left(0.4 \times 10^{-3}\right)^2\left(0.8 \times 10^{-3}\right)}=20 \times 10^{11} \mathrm{~N} / \mathrm{m}^2\end{aligned}$
Further $\quad \frac{\Delta Y}{Y}=2\left(\frac{\Delta d}{d}\right)+\left(\frac{\Delta l}{l}\right)$
$$
\begin{aligned}
\therefore \quad \Delta Y & =\left\{2\left(\frac{\Delta d}{d}\right)+\left(\frac{\Delta l}{l}\right)\right\} Y=\left\{2 \times \frac{0.01}{0.4}+\frac{0.05}{0.8}\right\} \times 2.0 \times 10^{11} \\
& =0.2 \times 10^{11} \mathrm{~N} / \mathrm{m}^2 \text { or }(Y+\Delta Y)=(2+0.2) \times 10^{11} \mathrm{~N} / \mathrm{m}^2
\end{aligned}
$$
The correct option is (b).

Topic: Physics, Mathematics in Physics, JEE Main
A student performs an experiment to determine young's modulus of a wire, exactly 2 m long, by searle's method. In a particular reading, the student measures the extension in the length of the wire to be 0.8 mm with an uncertainty of ± 0.05 mm at a load of exactly 1.0 kg. The student also measures the diameter of the wire to be 0.4 mm with an uncertainty of ± 0.01 mm. The young's modulus obtained from the reading is [Take g=9.8 ms-2]

Options:
A 2.0±0.3×1011 Nm-2
B 2.0±0.2×1011 Nm-2
C 2.0±0.1×1011 Nm-2
D 2.0±0.05×1011 Nm-2

The correct answer is:
2.0±0.2×1011 Nm-2

Solution:
Young's modulus Y=FLAl=4FLÏd2l
=41.0Ã9.82Ï0.4Ã10-320.8Ã10-3
=2.0Ã1011 Nm-2
Further, âYY=2âdd+âll
â´ âY=2âdd+âll Y
=2Ã0.010.4+0.050.8Ã2.0Ã1011
=0.2Ã1011 Nm-2
Or Y+âY=2+0.2Ã1011 Nm-2

Topic: Physics, Units and Dimensions, JEE Main
A student read common difference of an A. P. as

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

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

The correct answer is:
35

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

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

Topic: Mathematics, Sequences and Series, JEE Main
A student records the initial length $l$, change in temperature $\Delta T$ and change in length $\Delta l$ of a rod as follows
If the first observation is correct, what can you say about observation 2,3 and 4.

Options:

None

Solution:
As here in experiment is a rod and linear expansion $(\alpha)$ of a material or rod remain same i.e. equal to each ${ }^{\circ} \mathrm{C}$ as $L$ and $\Delta t$ is same.
From the 1st observation,
$$
\alpha=\frac{\Delta l}{l \Delta T} \Rightarrow \alpha=\frac{4 \times 10^{-4}}{2 \times 10}=2 \times 10^{-5 \circ} \mathrm{C}^{-1}
$$
$\Delta l$ value for 2 nd observation,
$$
\begin{aligned}
\Delta l &=\alpha l \Delta T=2 \times 10^{-5} \times 1 \times 10 \\
&=2 \times 10^{-4} \mathrm{~m} \neq 4 \times 10^{-4} \mathrm{~m}
\end{aligned}
$$
(Wrong)
$\Delta l$ value for 3 rd observation,
$$
\begin{aligned}
\Delta l &=\alpha l \Delta T=2 \times 10^{-5} \times 2 \times 20 \\
&=8 \times 10^{-4} \mathrm{~m} \neq 2 \times 10^{-4} \mathrm{~m}
\end{aligned}
$$
(Wrong)
$\Delta l$ value for 4th observation,
$$
\begin{aligned}
\Delta l &=\alpha l \Delta T=2 \times 10^{-5} \times 3 \times 10 \\
&=6 \times 10^{-4} \mathrm{~m}=6 \times 10^{-4} \mathrm{~m}
\end{aligned}
$$

Topic: Physics, Thermal Properties of Matter
A student says that he had applied a force F=-kx on a particle and the particle moved in simple harmonic motion. He refuses to tell whether k is constant or not. Assume that he has worked only with positive x and no other force acted on the particle

Options:
A as x increases k increases
B as x increases k decreases
C as x increases and k remains constant
D the motion cannot be simple harmonic

The correct answer is:
as x increases k increases

Solution:
It is given that F=-kx
But For SHM F=-k0x
Therefore by comparing them k=k0x
Therefore as x increases k increases

Topic: Physics, Oscillations, NEET
A student scores the following marks in five tests: 45,54,41,57,43. His score is not known for the sixth test. If the mean score is 48 in the six tests, then the standard deviation of the marks in six tests is:

Options:
A 103
B 1003
C 103
D 1003

The correct answer is:
103

Solution:
Given mean of 6 observations is 48

⇒ 41+45+54+57+43+x6=48

⇒x=48
Now variance σ2=∑x-x2n

=72+32+62+92+52+026
=1003
Hence standard deviation =103.

Topic: Mathematics, Statistics, JEE Main, JEE Main 2019 (08 Apr Shift 2)
A student uses the resistance of a known resistor (1 Ω ) to calibrate a voltmeter and an ammeter using the circuits shown below. The student measures the ratio of the voltage to current to be 1×103Ω in circuit
(a) and 0.999Ω in circuit (b). From these measurements, the resistance (in Ω ) of the voltmeter and ammeter are found to be close to

Options:
A 102 and 10-2
B 103 and 10-3
C 10-2 and 102
D 10-2 and 103

The correct answer is:
103 and 10-3

Solution:
When a voltmeter put in series, it still reads potential drop and when an ammeter is connected in parallel, it still shows current through it.
Case a

Let I= current through cell, then potential drop read by voltmeter is V=I·RV (this is reading of voltmeter) Where, RV is the resistance of voltmeter In loopA B, VAB=I1×1=I2×RA and I=I1+I2
Where, RA is the resistance of ammeter We substitute for I1 from above equation to get ⇒I=I2RA+I2=I2RA+1
⇒I2=IRA+1
(this is reading of ammeter)

Now given, voltmeter readin'g ammeter reading =1×103=IRVIRA+1
So,RVRA+1=10001000$
Case b

Let I= current through cell, then ammeter reading in this case is I.
Also, in loop A B,
VAB=I1×1=I2×RV

As, I=I1+I2=I2RV+I2

=I2RV+1
So, I2=IRV+1
Hence, voltmeter reading is V=I2RV =IRVRV+1 (this is reading of voltmeter)
Now given, voltmeter reading ÷ ammeter reading =0.999Ω.
So,0.999=IRVRV+1I

⇒0.999=RVRV+1

So,RV=999Ω
≈103Ω
Substituting RV in Eq (i), we get
RA=1999
or RA=10-3Ω

Topic: Physics, Current Electricity, JEE Main
A student while solving a quadratic equation in $x$, he copied its constant term incorrectly and got its roots as 5 and 9. Another student copied the constant term and coefficient of $x^2$ of the same equation correctly as 12 and 4 respectively. If $s, p$ and $\Delta$ denote respectively the sum of the roots, the product of the roots and the discriminate of the correct equation, then $\frac{\Delta}{3 p+s}=$

Options:
A 48
B 45
C 128
D 16

The correct answer is:
128

Solution:
Let the equation be $a x^2+b x+c=0$
Sum of roots $=-\frac{b}{a}$
and product of roots $=\frac{c}{a}$
when constant term is copied wrongly we get roots as 5 and 9.
$\therefore \quad-\frac{b}{a}=5+9 \quad$ [sum will remain correct]
$\Rightarrow \quad-\frac{b}{a}=14$
Another student copied constant term and coefficient of $x^2$ correctly us 12 and 4 .
$$
\begin{array}{ll}
\therefore & c=12 \text { and } a=4 \\
\therefore & \frac{-b}{a}=14 \Rightarrow b=-56 \\
\therefore & S=-\frac{b}{a}=\frac{-(-56)}{4}=14
\end{array}
$$
$$
\begin{aligned}
P & =\frac{c}{a}=\frac{12}{4}=3 \\
\Delta & =b^2-4 a c \\
& =(-56)^2-4 \times 4 \times 12 \\
& =3136-192=2944 \\
\therefore \frac{\Delta}{3 p+S} & =\frac{2944}{3 \times 3+14}=\frac{2944}{23}=128
\end{aligned}
$$

Topic: Mathematics, Quadratic Equation, TS EAMCET, TS EAMCET 2019 (04 May Shift 2)
A submarine A travelling at 18km hr-1 is being chased along the line of its velocity by another submarine B travelling at 27 km hr-1 . B sends a sonar signal of 500 Hz to detect A and receives a reflected sound of frequency v. The value of v is closed to (Speed of sound in water 1500 m s-1 )

Options:
A 504Â Hz
B 499Â Hz
C 502 Hz
D 507Â Hz

The correct answer is:
502 Hz

Solution:
VA=18 km h-1=5 m s-1 ; VB=27 km h-1=7.5m s-1
Frequency received by A, fA=f01500-51500-7.5
Frequency of reflected wave as heard by B.
f'B=fA1500+7.51500+5
=f01500-51500-7.51500+7.51500+5
=50014951492.51507.51505
=501.67 Hz≈502 Hz

Topic: Physics, Waves and Sound, JEE Main, JEE Main 2019 (12 Apr Shift 1)
A submarine experiences a pressure of 5.05×106 Pa at a depth of d1 in a sea. When it goes further to a depth of d2 , it experiences a pressure of 8.08×106 Pa . Then d2-d1 is approximately (density of water =103 kg/m3 and acceleration due to gravity =10 ms-2 ):

Options:
A 600 m
B 500 m
C 300 m
D 400 m

The correct answer is:
300 m

Solution:
The pressure at a depth d is p=p0+ρgd, here p0 is the atmospheric pressure.
∴p1=p0+ρgd1
and p2=p0+ρgd2
The difference in pressure is p2-p1=ρgd2-d1

Putting the values in the above equation
8.08×106-5.05×106=103×10d2-d1
d2-d1=3.03×106104
=303 m

Topic: Physics, Mechanical Properties of Fluids, JEE Main, JEE Main 2019 (10 Apr Shift 2)
A submarine has a window of area 30×30 cm2 on its ceiling and is at a depth of 100 m below sea level in a sea. If the pressure inside the submarine is maintained at the sea-level atmosphere pressure, then the force acting on the window is (consider density of sea water =1.03×103 kg m-3 acceleration due to gravity =10 m s-2):

Options:
A 0.93×105 N
B 0.93×103 N
C 1.86×105 N
D 1.86×103 N

The correct answer is:
0.93×105 N

Solution:
P=P0+ρgh

P= Pressure on upper surface of window =P0+ρgh

Pin= Pressure inside the submarine =P0

Net force =P0+ρghA-P0A

=ρghA

=1.03×103×10×100×900×10-4

=9.27×104 Newton

=0.93×105 Newton

Topic: Physics, Mechanical Properties of Fluids, KVPY, KVPY 2020 (SA)
A subset of 5 elements is chosen from the set of first 15 natural numbers. The probability that at least two of the five numbers are consecutive is Î», then the value of 22Î» is equal to

Options:

The correct answer is:
26

Solution:
Total cases =15C5
Unfavourable cases (no two are consecutive)=11C5
Required probability =1-11C515C5
=1-213=1113

Topic: Mathematics, Probability, JEE Main
A subspecies of plains zebra that lived in South Africa has recently become extinct is

Options:
A thylacine
B dodo
C quagga
D stellar sea cow

The correct answer is:
quagga

Solution:
Equus quagga quagga is a subspecies of plains zebra that lived in South Africa. It had brown and white stripes mainly in the front part of the body. The rear part of the body did not have stripes, so it looked like a horse. They were extensively hunted for their meat and skin and the wild species died in 1883.

Topic: Biology, Biodiversity and its Conservation, NEET