datavorous/entrance-exam-dataset huggingface dataset:
If $5 x^2+\lambda y^2=20$ represents a rectangular hyperbola, then $\lambda$ equals

Options:
A $5$
B $4$
C $-5$
D None of these

The correct answer is:
$-5$

Solution:
Since the general equation of second degree represents a rectangular hyperbola, if $\Delta \neq 0, h^2 \gt a b$ and coefficient of $x^2+$ coefficient of $y^2=0$. Therefore the given equation represents a rectangular hyperbola, if $\lambda+5=0$ i.e., $\lambda=-5$.

Topic: Mathematics, Hyperbola, JEE Main
If 5x+9=0 is the directrix of the hyperbola 16x2-9y2=144, then its corresponding focus is:

Options:
A -5, 0
B 5, 0
C -53, 0
D 53,Â 0

The correct answer is:
-5, 0

Solution:
The equation of a hyperbola in standard form is x2a2-y2b2=1 and its foci are ±ae, 0 and directrices are x=±ae.

The given equation of hyperbola is 16x2-9y2=144, ⇒x29-y216=1

⇒a2=9, b2=16

Given, directrix of the hyperbola is 5x+9=0, ⇒x=-95.

Hence, on comparing the directrix with the standard equation, we get ae=95

⇒3e=95

⇒e=53

And, hence the required focus is -ae, 0=(-5, 0).

Topic: Mathematics, Hyperbola, JEE Main, JEE Main 2019 (10 Apr Shift 2)
If 6312+10311+20310+4039+…..+102403=2n·m, where m is odd, then m.n is equal to _____ .

Options:

The correct answer is:
12

Solution:
Given,

6312+10311+20310+4039+…..+102403=2n·m

On rearranging we get,

⇒1312+520312+21311+22310+……+2113=2n·m

⇒1312+5131262-16-1=2n·m

⇒1312+551312·212·312-1312=2n·m

⇒1312+212-1312=2n·m

⇒2n·m=212

On comparing both side we get,

⇒ m=1 and n=12

So, m·n=12

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2022 (28 Jul Shift 2)
If $\left|\begin{array}{rrr}6 i & -3 i & 1 \\ 4 & 3 i & -1 \\ 20 & 3 & i\end{array}\right|=x+i y$ then $(x, y)$ is

Options:
A $(3,1)$
B $(1,3)$
C $(0,3)$
D $(0,0)$

The correct answer is:
$(0,0)$

Solution:
$\left|\begin{array}{rrr}6 i & -3 i & 1 \\ 4 & 3 i & -1 \\ 20 & 3 & i\end{array}\right|_{=} x+i y$
$\begin{aligned} & \left|\begin{array}{ccc}6 i+4 & 0 & 0 \\ 4 & 3 i & -1 \\ 20 & 3 & i\end{array}\right|=x+i y \quad\left[R_1 \rightarrow R_1+R_2\right] \\ \Rightarrow & (6 i+4)\left(3 i^2+3\right)=x+i y \\ \Rightarrow & (6 i+4)(-3+3)=x+i y \\ \Rightarrow & x+i y=0=0+i .0 \Rightarrow(x, y)=(0,0) .\end{aligned}$

Topic: Mathematics, Complex Number, JEE Main
If 8=3+143+p+1423+2p+1433+3p+…∞, then the value of p is

Options:

The correct answer is:
9

Solution:
Given that 3+143+p+1423+2p+…. to ∞=8

Let, S=3+143+p+1423+2p+...∞

⇒S= 3+34+342+...+p4+2p42+...

We know that, S∞=a+ar+ar2+...∞=a1-r

Let, S1=3+34+342+....

⇒S1=31+14+142+....

⇒S1=311-14=4

And, S2=p4+2p42+3p43+....

⇒S24=p42+2p43+3p44+....

⇒ S2-S24=p4+p42+p43+.....

⇒3S24=p14+142+143+.....

⇒3S24=p141-14=p3

⇒ S2=4p9

So, S=S1+S2=8

⇒ 4+4p9=8

⇒4p9=4

⇒p=9

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2024 (27 Jan Shift 1)
If $8 f(x)+6 f\left(\frac{1}{x}\right)=x+5$ and $y=x^2 f(x)$, then $\frac{d y}{d x}$ at $x=-1$ equals

Options:
A 0
B $\frac{1}{14}$
C $\frac{-1}{14}$
D 1

The correct answer is:
$\frac{-1}{14}$

Solution:
$\begin{aligned} 8 f(x)+6 f\left(\frac{1}{x}\right) & =x+5 \\ y & =x^2 f(x)\end{aligned}$
Replace $x$ by $\frac{1}{x}$ in Eq. (i), we obtain
$$
8 f\left(\frac{1}{x}\right)+6 f(x)=\frac{1}{x}+5
$$
Solve Eqs. (i) and (ii) for $f(x)$ and $f\left(\frac{1}{x}\right)$
Multiply by 8 in Eq. (i) and by 6 in Eq. (ii) and subtract, we get
$$
\begin{aligned}
& 64 f(x)-36 f(x)=8 x-\frac{6}{x}+10 \\
& \Rightarrow \quad 28 f(x)=8 x-\frac{6}{x}+10
\end{aligned}
$$
Now, $28 x^2 f(x)=8 x^3-6 x+10 x^2$
$$
\Rightarrow \quad 28 y=8 x^3-6 x+10 x^2
$$
Differentiating it,
$$
\Rightarrow \begin{aligned}
28 \frac{d y}{d x} & =24 x^2-6+20 x \\
\Rightarrow \quad \frac{d y}{d x} & =\frac{24 x^2-6+20 x}{28} \\
\left(\frac{d y}{d x}\right)_{x=-1} & =\frac{24(-1)^2-6+20(-1)}{28}=-\frac{1}{14}
\end{aligned}
$$

Topic: Mathematics, Differentiation, JEE Main
If ( 9a,6a ) is a point bounded in the region formed by parabola y 2 =16x and x=9, then

Options:
A a∈(0,1)
B a<14
C a<1
D 0<a<4

The correct answer is:
a∈(0,1)

Solution:
Since, the point ( 9a,6a ) is bounded in the region formed by the parabola y 2 =16 and x=9, then
y2−16x<0, x−9<0
⇒36a2−169a<0, 9a−9<0
36a(a−4)<0,  9a−9<0
0<a<4, a<1⇒0<a<1

Topic: Mathematics, Parabola, JEE Main
If (10)9+2(11)1(10)8+3(11)2(10)7+......+10(11)9 =k(10)9, then k is equal to :

Options:
A 100
B 110
C 1 2 1 1 0
D 4 4 1 1 0 0

The correct answer is:
100

Solution:
Let, S= ( 10 ) 9 +2 ( 11 ) 1 ( 10 ) 8 +3 ( 11 ) 2 ( 10 ) 7 +......+10 ( 11 ) 9 =k ( 10 ) 9
S=(10)91+21110+311102+.....+1011109
So,k=1+21110+311102+.....+1011109

1110k=1110+211102+311103+.....+10111010
Subtracting
−k10=1+1110+11102+.……+11109−10111010
Here,

First term =1
Common ratio = 11 10
Number of terms =10
∴ k10=10111010−1111010−11110−1
k=1010111010−10111010−1

k=100111010−100111010+100
k=100

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2014 (06 Apr)
If 1022 gas molecules each of mass 10-26 kg collides with a surface (perpendicular to it) elastically per second over an area 1 m2 with a speed 104m/s, the pressure exerted by the gas molecules will be of the order of:

Options:
A 2 Pa
B 4 Pa
C 6 Pa
D 8 Pa

The correct answer is:
2 Pa

Solution:
Total momentum transferred in one second =n2mv=10222×10-26×104=2
Force experienced by surface =2 N
Pressure on surface =ForceArea=21=2 Pa

Topic: Physics, Center of Mass Momentum and Collision, JEE Main, JEE Main 2019 (08 Apr Shift 1)
If 10 different balls are to be placed in 4 distinct boxes at random, then the probability that two of these boxes contain exactly 2 and 3 balls is:

Options:
A 965211
B 965210
C 945210
D 945211

The correct answer is:
945210

Solution:
Total number of ways in which 10 balls can be placed in 4 boxes =4×4×4×4×4×4×4×4×4×4

⇒410

⇒220

Exactly two box contain 2,3 balls

⇒C24×2×C210×C38×25

Probability =C24×2×C210×C28×25220

=945210

Topic: Mathematics, Probability, JEE Main, JEE Main 2020 (09 Jan Shift 2)
If 10% of the power of a 100π W light bulb is converted to visible radiation, then the average intensity of visible radiation at a distance of 10 m is

Options:
A 0.025 W m-2
B 0.01 W m-2
C 0.031 W m-2
D 0.05 W m-2

The correct answer is:
0.025 W m-2

Solution:
The intensity is defined as power radiated per unit area here power, P=100π W, but in the question we want to calculate only for ten percent of origin it means, P'=10π W, so the intensity,
I=P'4πd2=10π4π102=0.025 W m-2.

Topic: Physics, Electromagnetic Waves, JEE Main
If 11 arithmetic means are inserted between 28 and 10, then the number of integral arithmetic means are

Options:
A 5
B 6
C 7
D 8

The correct answer is:
5

Solution:
Let, A1,A2,A3,â¦â¦,A11 be 11 A.M.âs between 28 and 10
â28,A1,A2,â¦â¦,A11,10 are in A.P

Let, d be the common difference of A.P.
Also, the number of terms =13

â10=T13=T1+12d=28+12d
âd=10-2812=-1812=-32
â Number of integral A.M.âs are 5

Topic: Mathematics, Sequences and Series, JEE Main
If $1.2 \mathrm{~g}$ of metal displace 1.12 litre hydrogen at normal temperature and pressure ,equivalent weight of metal would be

Options:
A $24$
B $12$
C $1.2 \div 11.2$
D $1.2 \times 11.2$

The correct answer is:
$24$

Solution:
1.12ltr $\mathrm{H}_2 1.2 \mathrm{~g} ;, 22.4$ ltr $\mathrm{H}_2=24 \mathrm{~g}$.

Topic: Chemistry, Redox Reactions, JEE Main
If 12 identical balls are to be placed in 3 identical boxes, then the probability that one of the boxes contains exactly 3 balls is

Options:
A 22 13 11
B 519
C 55 2 3 1 0
D 220 1 3 1 2

The correct answer is:
519

Solution:
B1 B2 B3
12 0 0
11 1 0
10 1 1
10 2 0

9 3 0

9 2 1
8 4 0

8 3 1

8 2 2
7 5 0
7 4 1

7 3 2

6 6 0
6 5 1
6 4 2

6 3 3

5 5 2

5 4 3

4 4 4
Total number of possibilities=19.
Number of favourable case =5
Required probability =519

Topic: Mathematics, Probability, JEE Main, JEE Main 2015 (04 Apr)
If 1,2,p, 2,8,-6 and α2-2α,p,1 are ordered triplet pair of the form x,y,z which satisfy all the equations xa+yb+zc=1, xb+yc+za=1 and xc+ya+zb=1, then the sum of all the values of α is equal to (where, ab+bc+ca≠0)

Options:
A 3
B 2
C 0
D -2

The correct answer is:
2

Solution:
System of equation has more than one solution
∆=1a1b1c1b1c1a1c1a1b=0⇒1a=1b=1c⇒a=b=c
⇒ all three equations are x+y+z=a
Now, 2,8,-6 satisfy this equation
2+8-6=a⇒a=4
1,2,p also satisfy this equation
⇒1+2+p=4⇒p=1
α2-2α,p,1 also satisfy this equation
⇒α2-2α+1+1=4⇒α2-2α-2=0
Sum of the values of α =2

Topic: Mathematics, Determinants, JEE Main
If ${ }^{15} C_{3 r}={ }^{15} C_{r+3}$, then the value of $r$ is

Options:
A $3$
B $4$
C $5$
D $8$

The correct answer is:
$3$

Solution:
$15=3 r+r+3$
Hence, $r=3$

Topic: Mathematics, Permutation Combination, JEE Main
If 15sin4α+10cos4α=6, for some α∈R, then the value of 27sec6α+8cosec6α is equal to :

Options:
A 350
B 500
C 400
D 250

The correct answer is:
250

Solution:
15sin4α+10cos4α=6

15sin4α+10cos4α=6sin2α+cos2α2

3sin2α-2cos2α2=0

∴ 3sin2α-2cos2α=0

tan2α=23 & cot2α=32

⇒27sec6α+8cosec6α

=27sec2α3+8cosec2α3

=271+tan2α3+81+cot2α3

=271+233+81+323

=250

Topic: Mathematics, Trigonometric Ratios & Identities, JEE Main, JEE Main 2021 (18 Mar Shift 2)
If 17 arithmetic means are inserted between 72 and -832, then the 17th arithmetic mean is

Options:
A -19
B -29
C -39
D -49

The correct answer is:
-39

Solution:
Let d be the common difference. Then
d=-832-7218=-52
So, the 17th mean is given by
-832--52=-832+52=-782=-39

Topic: Mathematics, Sequences and Series, JEE Main
If 19 th term of a non-zero A.P. is zero, then its (49th term):

(29th term) is:

Options:
A 4: 1
B 1: 3
C 3: 1
D 2: 1

The correct answer is:
3: 1

Solution:
Let first term and common difference of $A P$ be a and d respectively, then

$t_{n}=a+(n-1) d$

$\therefore \quad t_{19}=a+18 d=0$

$\therefore \quad a=-18 d$

$\therefore \quad \frac{t_{49}}{t_{29}}=\frac{a+48 d}{a+28 d}$

$=\frac{-18 d+48 d}{-18 d+28 d}=\frac{30 d}{10 d}=3$

$t_{49}: t_{29}=3: 1$

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2019 (11 Jan Shift 2)
If 2019+2212018+32122017+...+202119=k2019, then k is equal to _____.

Options:

The correct answer is:
400

Solution:
Let,

S=2019+2212018+32122017+...+202119

⇒S=2019+2·2019·2120+3201921202+...+202019·212019

⇒S=20191+2·2120+321202+...+20212019

Now on comparing with S=k·2019 we get,

⇒ k=1+2·2120+321202+....+20212019 ....1

⇒2120k=2120+221202+...+19212019+20212020 ....2

Now on subtracting above equations we get,

⇒k-2120k=1+2120+21202+...+212019-20212020

⇒-k20=212020-12120-1-20212020

⇒-k20=212020-1×20-20212020

⇒-k20=-20

⇒k=400

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2023 (06 Apr Shift 2)
If â«2040sinâ¡xsinâ¡x+sin60-xdx=k, then the value of k4 is equal to

Options:

The correct answer is:
2.50

Solution:
As, k=∫2040sin⁡xsin⁡x+sin60-xdx .... (i)
Applying a+b-x property, we get,
k=∫2040sin⁡60-xsin60-x+sin⁡xdx ... (ii)
Adding (i) and (ii), we get,

2k=∫20401 dx=x2040
⇒2k=40-20
⇒k=10
∴k4=2.5

Topic: Mathematics, Definite Integration, JEE Main
If $20 \mathrm{~cm}^3$ gas at $1 \mathrm{~atm}$. is expanded to $50 \mathrm{~cm}^3$ at constant $T$, then what is the final pressure

Options:
A $20 \times \frac{1}{50}$
B $50 \times \frac{1}{20}$
C $1 \times \frac{1}{20} \times 50$
D None of these

The correct answer is:
$20 \times \frac{1}{50}$

Solution:
At constant $T, P_1 V_1=P_2 V_2$
$1 \times 20=P_2 \times 50 ; \quad P_2=\frac{20}{50} \times 1$

Topic: Chemistry, States of Matter, JEE Main
If 20 J of work has to be done to move an electric charge of 4 C from a point of potential 10 V to another point of potential V, then the value of V is

Options:
A 20 V
B 30 V
C 5 V
D 15 V

The correct answer is:
15 V

Solution:
The work done by the external agent is

Wext=qV2-V1

⇒20=4(V – 10)

⇒V=15 V

Topic: Physics, Electrostatics, JEE Main
If 27999 is divided by 7, then the remainder is

Options:
A 3
B 1
C 6
D 2

The correct answer is:
6

Solution:
Given, (27)999

=28-19997

=28λ-17

=28λ-7+7-17

=74λ-1+67

∴ Remainder=6

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2017 (08 Apr Online)
If 27999 is divided by 7, then the remainder is

Options:
A 1
B 2
C 3
D 6

The correct answer is:
6

Solution:
Express the given number 27 999 as 28−1 999
Now, divide by 7, (28−1) 999 7
and apply the formula of binomial expansion.
We get the reminder as −1⇒7+(−1)=6

Topic: Mathematics, Mathematical Induction, JEE Main
If 30 mL of H2 and 20 mL of O2 react to form water, what is left at the end of the reaction?

Options:
A 10 mL of H2
B 5 mL of H2
C 10 mL of O2
D 5 mL of O2

The correct answer is:
5 mL of O2

Solution:
2H2+O2â2H2O
Combining volume ratio of H2 and O2 =2:1
Here H2 is limiting reactant. 30 ml of H2 will react with 15 ml of O2 and 5 ml of O2 will remain unreacted at the end.

Topic: Chemistry, Some Basic Concepts of Chemistry, NEET
If $\frac{x^{2}}{36}-\frac{y^{2}}{k^{2}}=1$ is a hyperbola, then which of the following statements can be true?

Options:
A $(3,1)$ lies on the hyperbola.
B $(-3,1)$ lies on the hyperbola.
C $(5,2)$ lies on the hyperbola.
D $(10,4)$ lies on the hyperbola.

The correct answer is:
$(10,4)$ lies on the hyperbola.

Solution:
We have,
$$
\frac{x^{2}}{36}-\frac{y^{8}}{k^{2}}=1
$$
Option (a) $(3,1)$
$$
\begin{aligned}
\frac{9}{36}-\frac{1}{k^{2}} &=1 \\
\Rightarrow \quad & \frac{1}{k^{2}}=\frac{-27}{36} \Rightarrow k^{2}=\frac{-36}{27}
\end{aligned}
$$
which is not possible.
Option (b) $(-3,1)$
$$
\frac{9}{36}-\frac{1}{k^{2}}=1 \Rightarrow k^{2}=\frac{-36}{27}
$$
which is not possible.
Option (c) $(5,2)$
$$
\begin{aligned}
\frac{25}{36}-\frac{4}{k^{2}} &=1 \\
\frac{4}{k^{2}} &=\frac{-11}{36} \\
k^{2} &=\frac{-144}{11}
\end{aligned}
$$
which is not possible.
Option (d) $(10,4)$
$$
\frac{100}{36}-\frac{16}{k^{2}}=1 \Rightarrow \frac{16}{k^{2}}=\frac{64}{36}
$$
$\Rightarrow k^{2}=9$, which is possible.

Topic: Mathematics, Hyperbola, JEE Main
If 46 chromosomes are present at the G1 phase, then the number of chromosomes after S phase would be:

Options:
A 23
B 92
C 47
D 46

The correct answer is:
46

Solution:
The number of chromosome remains same in S phase, the amount of genetic material doubles.
In anaphase-telophase the number of chromosome becomes twice when the centromere splits.

Topic: Biology, Cell Cycle and Cell Division, NEET
If 50 mL of 0.5M oxalic acid is required to neutralise 25 mL of NaOH solution, the amount of NaOH in 50 mL of given NaOH solution is_______g.

Options:

The correct answer is:
4

Solution:
50ml of 0.5 M oxalic acid is completely neutralised by 25ml of NaOH solution.

For neutralisation reactions, N1V1=N2V2

Normality of oxalic acid = Molarity×2

⇒Noxalic acid=0.5×2=1N

50×1=25×NNaOH

For sodium hydroxide, molarity is the same as normality.

Molarity of sodium hydroxide = 2M

The number of moles of sodium hydroxide= 50×2×10-3 mol.

Hence, the mass of sodium hydroxide =50×2×10-3×40=4g

Topic: Chemistry, Redox Reactions, JEE Main, JEE Main 2024 (29 Jan Shift 2)
If 50 % of a reaction occurs in 100 second and 75 % of the reaction occurs in 200 second, the order of this reaction is:

Options:
A 2
B 3
C Zero
D 1

The correct answer is:
1

Solution:
50% of a reaction occurs in 100 second and 75% of the reaction occurs in 200 second, the order of this reaction is 1.

Let the initial reactant concentration is 1 M.

In the first 100 seconds, reactant concentration is reduced to 50% or 1×50100​=0.5 M.

In the second 100 seconds (total 200 seconds from start), reactant concentration is reduced to 100−75=25% or 1× 25100​ =0.25 M..

Thus, we observe that after each 100 second period, the reactant concentration is reduced to one half. Hence, 100 seconds is the half-life period and is independent of reactant concentration. This is characteristic of the first-order reaction.

First-order reaction as the half-life is constant.

Topic: Chemistry, Chemical Kinetics, JEE Main, JEE Main 2018 (16 Apr Online)
If 50 Vernier divisions are equal to 49 main scale divisions of a travelling microscope and one smallest reading of main scale is 0.5 mm the Vernier constant of travelling microscope is:

Options:
A 0.1 mm
B 0.1 cm
C 0.01 cm
D 0.01 mm

The correct answer is:
0.01 mm

Solution:
The number of divisions on the vernier scale =50

Main scale division (MSD)=0.5 mm

As, 50 divisions on the vernier scale coincide with 49 divisions on the main scale,

VSD=4950MSD

The minimum inaccuracy in the measurement of distance by a vernier scale,

= Least count of the vernier scale

=1 MSD–1 VSD

=1 MSD-4950MSD

=150MSD=150×0.5 mm

=0.01 mm.

Topic: Physics, Experimental Physics, JEE Main, JEE Main 2024 (30 Jan Shift 2)
If $60 \%$, of a first order reaction was completed in 60 minutes, $50 \%$ of the same reaction would be completed in approximately:
$$
(\log 4=0.60, \log 5=0.69)
$$

Options:
A 45 minutes
B 0 minutes
C 40 minutes
D 50 minutes

The correct answer is:
45 minutes

Solution:
For a first order reaction
$k=\frac{2.303}{t} \log \frac{a}{a-x}$
When $t=60$ and $x=60 \%$
$\begin{aligned}
k & =\frac{2.303}{60} \log \frac{100}{100-60} \\
& =\frac{2.303}{60} \log \frac{100}{40} \\
& =0.0153
\end{aligned}$
Now,
$\begin{aligned}
t^{1 / 2} & =\frac{2.303}{0.0153} \log \frac{100}{100-50} \\
& =\frac{2.303}{0.0153} \times \log 2 \\
& =\frac{2.303}{0.0153} \times 0.3010 \\
& =45.31 \mathrm{~min}
\end{aligned}$

Topic: Chemistry, Chemical Kinetics, JEE Main
If 75% of a first order reaction was completed in 90 minutes, 60% of the same reaction would be completed in approximately (in minutes) _____
(Take: log2=0.30 ;log2.5=0.40)

Options:

The correct answer is:
60

Solution:
T=2.303K log 100100−x%

T75%=2.303K log 10025=90

T60%=2.303K log 10040

T75%T60%=2log2log2.5 ⇒ 90T60%=2×0.30.4

T60%=90×46=60 min.

Topic: Chemistry, Chemical Kinetics, JEE Main, JEE Main 2020 (04 Sep Shift 1)
If $75 \%$ of a radioactive sample disintegrates in 16 days, the half-life of the radioactive sample is

Options:
A 6
B 4 days
C 8
D 12

The correct answer is:
8

Solution:
If $N_0$ be the initial amount of radioactive sample, then after $t=16$ day, remaining amount,
$\begin{aligned}
N & =N_0-75 \% \text { of } N_0 \\
& =N_0-\frac{75}{100} N_0=\frac{N_0}{4}
\end{aligned}$
Since, we know that,
$\begin{aligned}
& N=N_0\left(\frac{1}{2}\right)^n \Rightarrow \frac{N_0}{4}=N_0\left(\frac{1}{2}\right)^n \\
\Rightarrow & \frac{1}{4}=\left(\frac{1}{2}\right)^n \Rightarrow\left(\frac{1}{2}\right)^2=\left(\frac{1}{2}\right)^n \\
\Rightarrow & n=2 \Rightarrow \frac{t}{T_{1 / 2}}=2 \Rightarrow \frac{16}{T_{1 / 2}}=2 \\
\Rightarrow \quad & T_{1 / 2}=\frac{16}{2}=8 \text { days }
\end{aligned}$

Topic: Physics, Nuclear Physics, AP EAMCET, AP EAMCET 2020 (22 Sep Shift 2)
If 80 g of copper sulphate CuSO4·5H2O is dissolved in deionised water to make 5 L of solution. The concentration of the copper sulphate solution is x×10-3 mol L-1. The value of x is _____.

[Atomic masses Cu : 63.54u, S : 32u, O : 16u, H : 1u]

Options:

The correct answer is:
64

Solution:
Mole of CuSO4·5H2O=80249.5=0.32

Concentration of the copper sulphate solution = mole volume (L)=0.325=0.064 mol/L=64×10-3

Topic: Chemistry, Solutions, JEE Main, JEE Main 2021 (01 Sep Shift 2)
If $\alpha=0.98$ and current through emitter $i_c=20$ $m A$, the value of $\beta$ is

Options:
A 4.9
B 49
C 96
D 9.6

The correct answer is:
49

Solution:
$\beta=\frac{\alpha}{1-\alpha}=\frac{0.98}{1-0.98}=49$.

Topic: Physics, Semiconductors, JEE Main
If 100 mole of H2O2 decompose at 1 bar and 300 K, the work is done (kJ) by one mole of O2(g) as it expands against 1 bar pressure is:
2H2O2l⇌2H2Ol+O2(g)
R=8.3 J K-1 mol-1

Options:
A 124.50
B 249.00
C 498.00
D 62.25

The correct answer is:
124.50

Solution:
Irreversible process
W=P∆V=nRT
2H2O2100 moll⇌2H2Ol+O2   50 molg
∴ 100 mol H2O2 on decomposition will give 50 mol O2
=50×8.3×300
=124500 J
=124.50 kJ

Topic: Chemistry, Thermodynamics (C), JEE Main, JEE Main 2016 (10 Apr Online)
If $100$ times the $100^{\text {th }}$ term of an $AP$ with non zero common difference equals the $50$ times its $50^{\text {th }}$ term, then the $150^{\text {th }}$ term of this $A P$ is

Options:
A
$-150$

B
$150$ times its $50^{\text {th }}$ term

C
$150$

D
zero

The correct answer is:
zero

Solution:
$100\left(T_{100}\right)=50\left(T_{50}\right) \quad \Rightarrow 2[a+99 d]=a+49 d \Rightarrow a+149 d=0 \quad \Rightarrow T_{150}=0$

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2012 (Offline)
If $13.6 \mathrm{eV}$ energy is required to ionize the hydrogen atom, then the energy required to remove an electron from $n=2$ is

Options:
A
$10.2 \mathrm{eV}$

B
$0 \mathrm{eV}$

C
$3.4 \mathrm{eV}$

D
$6.8 \mathrm{eV}$

The correct answer is:
$3.4 \mathrm{eV}$

Solution:
$E_n=-\frac{13.6}{n^2} \Rightarrow E_2=-\frac{13.6}{2^2}=3.4 \mathrm{eV}$

Topic: Physics, Atomic Physics, JEE Main, JEE Main 2002
If $150 \mathrm{~J}$ of energy is incident on area $2 \mathrm{~m}^{2}$. If $Q_{r}=15 \mathrm{~J}$, coefficient of absorption is $0.6$, then amount of energy transmitted is

Options:
A $50 \mathrm{~J}$
B $45 \mathrm{~J}$
C $40 \mathrm{~J}$
D $30 \mathrm{~J}$

The correct answer is:
$45 \mathrm{~J}$

Solution:
When thermal radiations (Q) fall on a body, they are partly reflected, partly absorbed and partly transmitted.
$$
Q=Q_{a}+Q_{r}+Q_{t}
$$
and $\quad \frac{Q_{a}}{Q}+\frac{Q_{r}}{Q}+\frac{Q_{t}}{Q}=a+r+t=1$
$$
\Rightarrow \quad \frac{15}{150}+0.6+x=1
$$
or $0.1+0.6+x=1$
or $x=0.3$
Transmitting power, $t=\frac{Q_{t}}{Q}$
or $0.3=\frac{Q_{t}}{150}$
$$
\Rightarrow \quad Q_{t}=45 \mathrm{~J}
$$

Topic: Physics, Thermal Properties of Matter, JEE Main
If $150 \mathrm{~J}$ of heat is added to a system and the work done by the system is $110 \mathrm{~J}$, then change in internal energy will be

Options:
A $260 \mathrm{~J}$
B $150 \mathrm{~J}$
C $110 \mathrm{~J}$
D $40 \mathrm{~J}$

The correct answer is:
$40 \mathrm{~J}$

Solution:
$\Delta Q=\Delta U+\Delta W \Rightarrow \Delta U=\Delta Q-\Delta W=150-110=40 \mathrm{~J}$

Topic: Physics, Thermodynamics, JEE Main
If 200 MeV energy is released in the fission of a single nucleus of 92U235. How many fissions must occur per second to produce a power of 1 kW?

Options:
A 3.125 Ã 1013
B 6.250ÃÂ 1013
C 1.525Â Ã1013
D None of these

The correct answer is:
3.125 Ã 1013

Solution:
We know that 1 kW = 1 Ã 103 Js-1
Also, 1.6 Ã 10-9 J = 1 eV
â´200 MeV= 200 Ã 1.6 Ã 10-19 Ã 106 J
Number of fissions = PowerEnergy released
= 103200 Ã 1.6 Ã10-13= 3.125 Ã 1013

Topic: Physics, Nuclear Physics, JEE Main
If $200 \mathrm{MeV}$ of energy is released in the fission of one nucleus of ${ }_{92}^{236} \mathrm{U}$, the number of nuclei that must undergo fission to release an energy of $1000 \mathrm{~J}$ is

Options:
A $3.125 \times 10^{13}$
B $6.25 \times 10^{13}$
C $12.5 \times 10^{13}$
D $3.125 \times 10^{14}$

The correct answer is:
$3.125 \times 10^{13}$

Solution:
Energy released in the fission of one nucleus
$\begin{aligned}
& =200 \mathrm{MeV} \\
& =200 \times 1.6 \times 10^{-13} \mathrm{~J} \\
& =3.2 \times 10^{-11} \mathrm{~J}
\end{aligned}$
Energy to be released
$=1000 \mathrm{~J}$
$3.2 \times 10^{-11} \mathrm{~J}$ energy is released due to number of nucleus fissioned $=1$
$\therefore 1000 \mathrm{~J}$ energy is released due to number of nucleus fissioned
$\begin{aligned}
& =\frac{1 \times 1000}{3.2 \times 10^{-11}} \\
& =3.125 \times 10^{13} \text { nucleus }
\end{aligned}$

Topic: Physics, Nuclear Physics, JEE Main
If 200 mL of a 0.031 molar solution of H 2 SO 4 are added to 84 mL of a 0.150 M KOH solution, what is the pH of the resulting solution?

Options:
A 12.4
B 1.7
C 2.2
D 10.85

The correct answer is:
10.85

Solution:
mmol of H+initial=200Ã0.031Ã2=12.4
mmol of OH-initial=84Ã0.15=12.6
mmol of OH-left after neutralisation=0.2
OH − final  =  0.2 284  = 7 ×  10 −4  M
pOHâ=â3.15 and pHâ=â10.85

Topic: Chemistry, Ionic Equilibrium, JEE Main
If 250cm3of an aqueous solution containing 0.73g of a protein A is isotonic with one litre of another aqueous solution containing 1.65g of a protein B, at 298K, the ratio of the molecular masses of A and B is ___________×10−2 (to the nearest integer).

Options:

The correct answer is:
177

Solution:
For isotonic solution

i1C1=i2C2 {For protein i=1}

C1=C2

⇒ 0.73×1000MA×250=1.65MB×1

MAMB=0.73×41.65=1.77=177×10-2

Topic: Chemistry, Solutions, JEE Main, JEE Main 2020 (03 Sep Shift 2)