datavorous/entrance-exam-dataset huggingface dataset:
The relationship between the force F and position x of a body is as shown in the figure. The work done in displacing the body from x=1 mto x=5 m will be

Options:
A 30 J

B 15 J

C 25 J

D 20 J

The correct answer is:
20 J

Solution:
Work done = force x displacement
= area of ABNM + area of CDEN + area of EFGH + area of HIJ - area of OAM
=-12×1×10+1×5×5+12×1×10=20 J

Topic: Physics, Work Power Energy, JEE Main
The relationship between Young's modulus $Y$, Bulk modulus $\mathrm{K}$ and modulus of rigidity $\eta$ is

Options:
A $Y=\frac{9 \eta K}{\eta+3 K}$
B $\frac{9 Y K}{Y+3 K}$
C $Y=\frac{9 \eta K}{3+K}$
D $Y=\frac{3 \eta K}{9 \eta+K}$

The correct answer is:
$Y=\frac{9 \eta K}{\eta+3 K}$

Solution:
$Y=3 K(1-2 \sigma)$ and $Y=2 n(1+\sigma)$
Eliminating $\sigma$ we get $Y=\frac{9 \eta K}{\eta+3 K}$

Topic: Physics, Mechanical Properties of Solids, JEE Main
The relative error in the determination of the surface area of a sphere is \( \alpha \). The relative error in the determination of its volume is

Options:
A \( \frac{3}{2} \alpha \)
B \( \frac{2}{3} \alpha \)
C \( \alpha \)
D \( \frac{5}{2} \alpha \)

The correct answer is:
\( \frac{3}{2} \alpha \)

Solution:
Δss=2×Δrr⇒ΔVV=3×Δrr
∴   ΔVV  =  32  ΔssΔVV  =  32 α

Topic: Physics, Mathematics in Physics, JEE Main
The relative error in the determination of the surface area of a sphere is α . The relative error in the determination of its volume is

Options:
A 32α
B 23α
C α
D 52α

The correct answer is:
32α

Solution:
Δss=2×Δrr⇒ΔVV=3×Δrr
∴   ΔVV  =  32  ΔssΔVV  =  32 α

Topic: Physics, Mathematics in Physics, JEE Main, JEE Main 2018 (15 Apr)
The relative error in the determination of the surface area of a sphere is $\alpha$. Then the relative error in the determination of its volume is

Options:
A
$\frac{2}{3} \alpha$

B
$\frac{2}{3} \alpha$

C
$\frac{3}{2} \alpha$

D
$\alpha$

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

Solution:
Relative error in Surface area, $\frac{\Delta \mathrm{s}}{\mathrm{s}}=2 \times \frac{\Delta \mathrm{r}}{\mathrm{r}}=\alpha$ and relative error in volume, $\frac{\Delta \mathrm{v}}{\mathrm{v}}=3 \times \frac{\Delta \mathrm{r}}{\mathrm{r}}$
$\therefore$ Relative error in volume w.r.t. relative error in area,
$$
\frac{\Delta \mathrm{v}}{\mathrm{v}}=\frac{3}{2} \alpha
$$

Topic: Physics, Mathematics in Physics, JEE Main, JEE Main 2018 (15 Apr Shift 1 Online)
The relative humidity of air is 80% at 27   o C. .If the aqueous tension at the same temperature is 27 mm of Hg. The partial pressure of water vapour in the air will be (in mm of Hg ).

Options:

The correct answer is:
21.6

Solution:
Relative humidity =Partial pressureVapour pressureÃ100
âPartial pressure27Ã100=80
âPartial pressure=80Ã27100
â´Partial pressure=21.6 mm of Hg

Topic: Chemistry, States of Matter, JEE Main
The relative lowering of the vapour pressure of an aqueous solution containing a nonvolatile solute is 0.0125. The molality of the solution is

Options:
A 0.70 m
B 0.50 m
C 0.80 m
D 0.40 m

The correct answer is:
0.70 m

Solution:
As we know
po-ppo=x1= mole fraction of solute
The ratio po-ppo is the relative lowering of vapour pressure, which is equal to 0.0125 here.
So X1=0.0125
The relation between the mole fraction and molality is
1X1-1=1000m×18 (molecular weight of H2O=18 )
or 10.0125-1=100m×18 w
m=0.70 m

Topic: Chemistry, Solutions, JEE Main
The relative order of electronegativity of C,Ge, and Pb is

Options:
A C>Ge>Pb
B Ge>C>Pb
C Pb>C>Ge
D C>Pb>Ge

The correct answer is:
C>Pb>Ge

Solution:
In the carbon family, the electronegativity decreases from carbon to silicon but remains constant from silicon to tin, then slightly increases for lead. And the elements of the carbon family are more electronegative than that of boron family due to the small atomic size.

So, correct order of electronegativity of given elements will be

C > Pb>Ge

Topic: Chemistry, Classification of Elements and Periodicity in Properties, TS EAMCET, TS EAMCET 2022 (18 Jul Shift 1)
The relative reactivities of acyl compounds towards nucleophilic substitution are in the order of

Options:
A Acyl chloride $>$ Ester $>$ Acid anhydride $>$ Amide
B Acyl chloride $>$ Acid anhydride $>$ Ester $>$ Amide
C Ester $>$ Acyl chloride $>$ Amide $>$ Acid anhydride
D Acid anhydride $>$ Amide $>$ Ester $>$ Acyl chloride

The correct answer is:
Acyl chloride $>$ Acid anhydride $>$ Ester $>$ Amide

Solution:
Better is the leaving group higher will be the reactivity of acyl compounds towards nucleophilic acyl substitution. Weaker is the base, better is the leaving group. Weaker bases are derived from stronger acids.

Topic: Chemistry, Hydrocarbons, NEET, NEET 2008 (Mains)
The relative stability of +1 oxidation state of group 13 elements follows the order

Options:
A $\mathrm{Al} < \mathrm{Ga} < \mathrm{Tl} < \mathrm{In}$
B $\mathrm{T} 1 < \mathrm{In} < \mathrm{Ga} < \mathrm{Al}$
C $\mathrm{Ga} < \mathrm{Al} < \ln < \mathrm{Tl}$
D $\mathrm{Al} < \mathrm{Ga} < \mathrm{In} < \mathrm{Tl}$

The correct answer is:
$\mathrm{Al} < \mathrm{Ga} < \mathrm{In} < \mathrm{Tl}$

Solution:
Due to inert pair effect, the stability of +1 oxidation

state increases down the group. Thus, correct order of stability is $\mathrm{Al} < \mathrm{Ga} < \mathrm{In} < \mathrm{Tl}$

Topic: Chemistry, p Block Elements (Group 13 & 14), JEE Main, JEE Main 2019 (11 Jan Shift 2)
The relative uncertainty in the period of a satellite orbiting around the earth is 10-2 . If the relative uncertainty in the radius of the orbit is negligible, the relative uncertainty in the mass of the earth is:

Options:
A 2 × 10-2
B 6 × 10-2
C 3 × 10-2
D 10-2

The correct answer is:
2 × 10-2

Solution:
From Kepler's law,
T2=4π2GMr3
∆MM=2∆TT=2 ×10-2

Topic: Physics, Gravitation, JEE Main, JEE Main 2018 (16 Apr Online)
The relatively high boiling point of $H F$ is due to

Options:
A Hydrogen bonding
B Covalent bonding
C Unshared electron pair on $F$
D Being a halogen acid

The correct answer is:
Hydrogen bonding

Solution:
Due to the high partial charges on $\mathrm{H}$ and $\mathrm{F}$ in $\mathrm{HF}$, a Hydrogen bond between $\mathrm{HF}$ molecules is much stronger than that between NH3 molecules. The high boiling point of $\mathrm{HF}$ is thus due to the strong $\mathrm{H}$-bond.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
The relaxation time in conductors

Options:
A Increases with the increase of temperature
B Decreases with the increase of temperature
C It does not depend on temperature
D All of sudden changes at $400 \mathrm{~K}$

The correct answer is:
Decreases with the increase of temperature

Solution:
Because as temperature increases, the resistivity increases and hence the relaxation time decreases for conductors $\left(r \propto \frac{1}{\rho}\right)$.

Topic: Physics, Current Electricity, JEE Main
The remainder and the quotient of the binary division $\begin{array}{ll}(101110)_{2} \div(110), \text { are respectively } & \end{array}$

Options:
A $(111)_{2}$ and $(100)_{2}$
B $(100)_{2}$ and $(111)_{2}$
C $(101)_{2}$ and $(101)_{2}$
D $(100)_{2}$ and $(100)$.

The correct answer is:
$(100)_{2}$ and $(111)_{2}$

Solution:
$(101110)_{2}=1 \times 2^{5}+0 \times 2^{4}+1 \times 2^{3}+1 \times 2^{2}+1 \times 2^{1}+0 \times 2^{0}$
$=32+0+8+4+2+0$
$=(46)_{10}$
Similarly, $(110)_{2}=1 \times 2^{2}+1 \times 2^{1}+0 \times 2^{4}$
$=4+2$
$=(6)_{10}$
Quotient $=7$ Remainder $=4$ $(7)_{10}=(111)_{2}$ and $(4)_{10}=(100)_{2}$

Topic: Mathematics, Sets and Relations, NDA, NDA 2017 (Phase 2)
The remainder left out when $8^{2 n}-(62)^{2 n+1}$ is divided by 9 is

Options:
A
0

B
2

C
7

D
8

The correct answer is:
2

Solution:
$$
\begin{aligned}
& 8^{2 n}-(62)^{2 n+1}=(1+63)^n-(63-1)^{2 n+1} \\
& =(1+63)^n+(1-63)^{2 n+1}=\left(1+{ }^n c_1 63+{ }^n c_2(63)^2+\ldots .+(63)^n\right)+\left(1-{ }^{(2 n+1)} c_1 63+{ }^{(2 n+1)} c_2(63)^2+\ldots .+(-1)(63)^{(2 n+1)}\right) \\
& =2+63\left({ }^n c_1+{ }^n c_2(63)+\ldots .+(63)^{n-1}-{ }^{(2 n+1)} c_1+{ }^{(2 n+1)} c_2(63)+\ldots .(63)^{(2 n)}\right)
\end{aligned}
$$
$\therefore$ Reminder is 2

Topic: Mathematics, Binomial Theorem, JEE Main
The remainder left out when $8^{2 n}-(62)^{2 n+1}$ is divided by 9 is

Options:
A
0

B
2

C
7

D
8

The correct answer is:
2

Solution:
$$
\begin{aligned}
& 8^{2 n}-(62)^{2 n+1}=(1+63)^n-(63-1)^{2 n+1} \\
& =(1+63)^n+(1-63)^{2 n+1}=\left(1+{ }^n c_1 63+{ }^n c_2(63)^2+\ldots .+(63)^n\right)+\left(1-{ }^{(2 n+1)} c_1 63+{ }^{(2 n+1)} c_2(63)^2+\ldots .+(-1)(63)^{(2 n+1)}\right) \\
& =2+63\left({ }^n c_1+{ }^n c_2(63)+\ldots .+(63)^{n-1}-{ }^{(2 n+1)} c_1+{ }^{(2 n+1)} c_2(63)+\ldots .(63)^{(2 n)}\right)
\end{aligned}
$$
$\therefore$ Reminder is 2

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2009
The remainder obtained when $(1 !)^{2}+(2 !)^{2}+(3 !)^{2}+\ldots+(100 !)^{2}$ is divided by $10^{2}$ is

Options:
A 14
B 17
C 28
D 27

The correct answer is:
17

Solution:
Here, terms greater than $5 !$, i.e. $(5 !)^{2},(6 !)^{2}, \ldots \ldots,(100 !)^{2}$ is divisible by 100 .
$\therefore$ For terms
$(5 !)^{2},(6 !)^{2}, \ldots(100 !)^{2}$ remainder is 0 .
Now, consider $(1 !)^{2}+(2 !)^{2}+(3 !)^{2}+(4 !)^{2}$
$=1+4+36+576$
$=617$
When, 617 is divided by 100 , its remainder is 17 . So, required remainder is 17 .

Topic: Mathematics, Mathematical Induction, COMEDK, COMEDK 2016
The remainder obtained when $1 !+2 !+3 !+\ldots+11 !$ is divided by 12 is

Options:
A 9
B 8
C 7
D 6

The correct answer is:
9

Solution:
Let $S=1!+2 !+3 !+41+\ldots+11 !$
Here, we see that from $4 !$ to $111,$ we alwars get 12 factor, so it is always divisible by 12
Now. $1 !+2 !+3 !=1+2+6=9$
Hence, when $S$ is divided by $12,$ the remainder is 9.

Topic: Mathematics, Mathematical Induction, WBJEE, WBJEE 2014
The remainder obtained when \( 1 !+2 !+3 !+\ldots+11 ! \) is divided by \( 12 \) is

Options:
A \( 09 \)
B \( 08 \)
C \( 7 \)
D \( 06 \)

The correct answer is:
\( 09 \)

Solution:
Given that, $1 !+2 !+3 !+\cdot s+11 !$
Since $4 !=4 \times 3 \times 2 \times 1$ is divisible by 12
$5 !=5 \times 4 \times 3 \times 2 \times 1$ is divisible by 12
And so on.
$11 !=11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1$
is divisible by 12
So, reminder is $1 !+2 !+3 !=1+2+6=9$

Topic: Mathematics, Application of Derivatives, KCET, KCET 2015
The remainder obtained when $5^{99}$ is divided by 13 is

Options:
A 6
B 7
C 8
D 9

The correct answer is:
8

Solution:
No solution. Refer to answer key.

Topic: Mathematics, Basic of Mathematics, AP EAMCET, AP EAMCET 2017 (25 Apr Shift 1)
The remainder obtained when $5^{124}$ is divided by 124 is

Options:
A 5
B 0
C 2
D 1

The correct answer is:
5

Solution:
We have, $5^{124}=\left(5^{3}\right)^{41} \cdot 5$
Now, $\quad 5^{3}=1(\bmod 124)$
$\therefore \quad\left(5^{3}\right)^{41}=1(\bmod 124)$
$\left(5^{3}\right)^{41} \cdot 5=1 \cdot 5(\bmod 124)$
$\Rightarrow \quad 5^{124}=5(\bmod 124)$

Topic: Mathematics, Binomial Theorem, KCET, KCET 2007
The remainder obtained when $5^{124}$ is divided by 124 is

Options:
A 0
B 5
C 1
D 2

The correct answer is:
5

Solution:
We have;
$$
\begin{aligned}
5^{124} &=\left(5^{3}\right)^{41} \cdot 5 \\
\text { Now, } 5^{3} &=1 \cdot(\bmod 124) \\
\left(5^{3}\right)^{41} &=1\langle\bmod 124) \\
\left(5^{3}\right)^{41} \cdot 5 &=1 \times 5(\bmod 124)=5(\bmod 124)
\end{aligned}
$$
So, the remainder obtained when $5^{124}$ is divided by 124 is 5 .

Topic: Mathematics, Binomial Theorem, COMEDK, COMEDK 2020
The remainder obtained when 7100 is divided by 13 is

Options:
A 3
B 7
C 9
D 11

The correct answer is:
9

Solution:
7100=7250=4950
=52-350
=C0505250-C1505249⋅3+...-C495052349+C5050350
=13K+350=13K+3316⋅32
=13K+26+116⋅9
=13K+9C016⋅2616+C116⋅2615+...+C1516⋅26+C1616
=13K+913K1+1
=13K+9K1+9

Topic: Mathematics, Binomial Theorem, JEE Main
The remainder obtained when $11+2 !+95 !$ is divided by 15 is

Options:
A 14
B 3
C 1
D 0

The correct answer is:
3

Solution:
since, all terms from 5 lonwards are divisible by 15 and $1!+2!+3!+4 !=33$
The required remainder after dividing by 15 will be 3 .

Topic: Mathematics, Mathematical Induction, WBJEE, WBJEE 2012
The remainder obtained when 2740 is divided by 12 is

Options:

The correct answer is:
9

Solution:
2740=3340=3120
3119=4-1119
=C01194119-C11194118+C21194117-.…….+C1181194-C119119
=4k-1
⇒3120=34k-1=12k-3
=12k-1+9
So, the required remainder is 9.

Topic: Mathematics, Binomial Theorem, JEE Main
The remainder of $n^4-2 n^3-n^2+2 n-26$ when divided by 24 , is

Options:
A 20
B 21
C 22
D 23

The correct answer is:
22

Solution:
Let $\begin{aligned} f(n) & =n^4-2 n^3-n^2+2 n-26 \\ & =n^3(n-2)-n(n-2)-26 \\ & =(n-2)\left(n^3-n\right)-26 \\ & =(n-2) n\left(n^2-1\right)-26 \\ & =(n-2)(n-1) n(n+1)-26\end{aligned}$
$$
=24 k-48+22
$$
$[\because$ product of four consecutive natural numbers is divisible by 24]
$$
=24[k-2]+22
$$
Hence, remainder is 22.

Topic: Mathematics, Quadratic Equation, TS EAMCET, TS EAMCET 2015
The remainder on dividing 1+3+32+33+…+32021 by 50 is _____.

Options:

The correct answer is:
4

Solution:
Let S=1+3+32+⋯·32021

=32022-13-1=32022-12

Simplifying by using binomial expansion we get321011-12=(10-1)1011-12

⇒ 1011C0101011-1011C1101010+…+C10101011101-C10111011×1-12

100q+1011×10-22

⇒ 100q+10110-22=100q+101082=50q+5054

Now dividing 50q+5054 by 50

We get 4 as remainder.

Hence option B is correct.

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2022 (24 Jun Shift 2)
The remainder when, $10^{10} \cdot\left(10^{10}+1\right)\left(10^{10}+2\right)$ is divided by 6 is

Options:
A 2
B 4
C 0
D 6

The correct answer is:
0

Solution:
Here, $10^{10} \cdot\left(10^{10}+1\right) \cdot\left(10^{10}+2\right)$ is a product of 3 consecutive integers.
Therefore, it is divisible by 3 ! i.e., 6 .
So, required remainder is 0 .

Topic: Mathematics, Binomial Theorem, KCET, KCET 2013
The remainder when 22003 is divided by 17 is

Options:
A 1
B 2
C 8
D 11

The correct answer is:
8

Solution:
22003=2322000=824500
=816500=817-1500
=8C5000⋅17500-C5001⋅17499+.……..-C500499⋅17+C500500
=817k+1=8×17k+8
So, the remainder obtained is 8.

Topic: Mathematics, Binomial Theorem, JEE Main
The remainder when $3^{100} \times 2^{50}$ is divided by 5 is

Options:
A 1
B 2
C 3
D 4

The correct answer is:
4

Solution:
Now, $\quad 3^{2} \equiv 4(\bmod 5)$
$\Rightarrow \quad\left(3^{2}\right)^{2} \equiv 16(\bmod 5) \equiv 1(\bmod 5)$
$\Rightarrow \quad\left(3^{4}\right)^{25} \equiv(1)^{25}(\bmod 5)$
$\Rightarrow \quad 3^{100} \equiv 1(\bmod 5)$
and $\quad 2^{2} \equiv 4(\bmod 5)$
$\Rightarrow \quad\left(2^{2}\right)^{2} \equiv 16(\bmod 5) \equiv 1(\bmod 5)$
$\Rightarrow \quad\left(2^{4}\right)^{12} \equiv 1^{12}(\bmod 5) \equiv 1(\bmod 5)$
$\Rightarrow \quad 2^{48} \cdot 2^{2} \equiv 4(\bmod 5)$
$\therefore \quad 3^{100} \times 2^{50}=1 \times 4(\bmod 5)$
Hence, remainder is 4 .

Topic: Mathematics, Binomial Theorem, KCET, KCET 2008
The remainder when $3^{100} \times 2^{50}$ is divided by 5 is

Options:
A 1
B 2
C 3
D 4

The correct answer is:
4

Solution:
We have,
$$
\begin{aligned}
3^{2} &=4(\bmod 5) \\
\left(3^{2}\right)^{2} &=16(\bmod 5)=1(\bmod 5) \\
\left(3^{4}\right)^{25} &=(1)^{26}(\bmod 5) \\
\Rightarrow \quad 3^{100} &=1(\bmod 5) \\
\text { Again } 2^{2} &=4(\bmod 5) \\
\left(2^{2}\right)^{2} &=16(\bmod 5)=1(\bmod 5) \\
\left(2^{4}\right)^{12} &=(1)^{12}(\bmod 5)=1(\bmod 5) \\
2^{18} \cdot 2^{2} &=4(\bmod 5) \\
\therefore 3^{100} \times 2^{50} &=1 \times 4(\bmod 5)=4(\bmod 5)
\end{aligned}
$$

Topic: Mathematics, Binomial Theorem, COMEDK, COMEDK 2019
The remainder when 32022 is divided by 5 is

Options:
A 1
B 2
C 3
D 4

The correct answer is:
4

Solution:
321011 =10-11011 =-1-101011

=-1-C1011110+C21011102…=10q-1

So, here the remainder will be 5-1=4

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2022 (24 Jun Shift 1)
The remainder when $7^{7^{7^{...7}}}(22$ times 7$)$ is divided by 48 is

Options:
A 21
B 7
C 47
D 1

The correct answers are:
7

Solution:
$7^{7^{7-7}}=7^{2 \mathrm{~m}+1}=7(49)^{\mathrm{m}}=7(1+48)^{\mathrm{m}}=48 \mathrm{k}+7$

Topic: Mathematics, Binomial Theorem, WBJEE, WBJEE 2021
The remainder when $7^n-6 n-50(n \in N)$ is divided by 36 , is

Options:
A 22
B 23
C 1
D 21

The correct answer is:
23

Solution:
We have, $7^n=(1+6)^n$
$$
\begin{aligned}
& ={ }^n C_0+{ }^n C_1 6{ }^1+{ }^n C_2 6^2+{ }^n C_3 6^3+\ldots+{ }^n C_n 6^n \\
& =1+6 n+66^2\left[{ }^n C_2+{ }^n C_3 6+\ldots+{ }^n C_n 66^{n-2}\right] \\
& =1+6 n+36 \lambda \quad\left[\text { where, }{ }^n C_2+\ldots+{ }^n C_n 6^{n-2}=\lambda\right] \\
& \Rightarrow \quad 7^n-6 n=36 \lambda+1 \\
& \Rightarrow \quad 7^n-6 n-50=36 \lambda-49 \\
& \Rightarrow \quad 7^n-6 n-50=36 \lambda-72+23 \\
& \Rightarrow \quad 7^n-6 n-50=36(\lambda-2)+23
\end{aligned}
$$
$$
\Rightarrow \quad 7^n-6 n-50=36 \mu+23[\text { where } \lambda-2=\mu]
$$
$\therefore$ When $7^n-6 n-50$ is divided by 36 , then remainder will be equal to 23 .

Topic: Mathematics, Binomial Theorem, TS EAMCET, TS EAMCET 2016
The remainder when 19200+23200 is divided by 49, is _____ .

Options:

The correct answer is:
29

Solution:
Solving 19200+23200=21-2200+21+2200

=2C020021200×20+C220021198×22....C200200210×2200

Now we know that 212=441 is divisible by 49,

So rest of the terms will be divisible by 49

So lets consider for 2C200200210×2200=2201

Now rewriting the term 2201=2367=7+167

Now 1+767=C067×1+C1677+C26772.........

=1+67×7+49k

Now when dividing 1+767 by 49 we consider 1+67×7 as 49k is divisible by 49

Now 1+67×7=470 which when divided by 49 leaves remainder as 29.

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2023 (01 Feb Shift 1)
The remainder when 20212022+20222021 is divided by 7 is

Options:
A 0
B 1
C 2
D 6

The correct answer is:
0

Solution:
We know 2023=7×289

So 20212022+20222021

=2023-22022+2023-12021

=7m1+22022+7m2+-12021

=7m1+m2+8674-1

=7m1+m2+(7+1)674-1

=7m1+m2+7m3+1-1

=7m1+m2+m3

This is a multiple of 7

Hence, remainder is 0

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2022 (27 Jul Shift 1)
The remainder when 20212023 is divided by 7 is

Options:
A 2
B 3
C 4
D 5

The correct answer is:
5

Solution:
The remainder when 2021 divided by 7 is -2. Hence, the problem reduces to finding the remainder when -22023 is divided by 7.

=-22022×-27

=-2×220227

=-2×236747

=-2×86747

=-2×1+76747

Using binomial theorem, we get

=-2×1+7k7 where k is an integer.

=-2-14k7

Clearly, the remainder when -14k is divisible by 7 is 0.

=-27

=-2-5+57=57

Hence, the remainder is 5.

Topic: Mathematics, Binomial Theorem, JEE Main
The remainder when 20212023 is divided by 7 is

Options:
A 2
B 3
C 4
D 5

The correct answer is:
5

Solution:
The remainder when 2021 divided by 7 is -2. Hence, the problem reduces to finding the remainder when -22023 is divided by 7.

=-22022×-27

=-2×220227

=-2×236747

=-2×86747

=-2×1+76747

Using binomial theorem, we get

=-2×1+7k7 where k is an integer.

=-2-14k7

Clearly, the remainder when -14k is divisible by 7 is 0.

=-27

=-2-5+57=57

Hence, the remainder is 5.

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2022 (26 Jun Shift 1)
The remainder when the determinant
$$
\left|\begin{array}{lll}
2014^{2014} & 2015^{2015} & 2016^{2016} \\
2017^{2017} & 2018^{2018} & 2019^{2019} \\
2020^{2020} & 2021^{2021} & 2022^{2022}
\end{array}\right|
$$
is divided by 5 is.

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

The correct answer is:
4

Solution:
$\left|\begin{array}{ccc}(2015-1)^{2014} & (2015)^{2015} & (2015+1)^{2016} \\ (2017)^{2017} & (2018)^{2018} & (2019)^{2019} \\ (2020)^{2020} & (2021)^{2021} & (2022)^{2022}\end{array}\right|$
Raminder $=\left|\begin{array}{ccc}1 & 0 & 1 \\ 2^{2017} & 2^{2018} & -1 \\ 0 & 1 & 2^{2022}\end{array}\right|$
$=\left(4^{4040}+1\right)+\left(2^{2017}\right)$
$1+(5-1)^{4000}+2(4)^{1008}=1+1+2(1)=4$

Topic: Mathematics, Determinants, KVPY, KVPY 2015 (SB/SX)
The remainder when the polynomial $1+x^{2}+x^{4}+x^{6}+\ldots+x^{22}$ is divided by $1+x+x^{2}+x^{3}+\ldots .+x^{11}$ is -

Options:
A 0
B 2
C $1+x^{2}+x^{4}+\ldots+x^{10}$
D $2\left(1+x^{2}+x^{4}+\ldots+x^{10}\right)$

The correct answer is:
$2\left(1+x^{2}+x^{4}+\ldots+x^{10}\right)$

Solution:
$$
\begin{array}{l}
P(x)=1+x^{2}+x^{4}+x^{6}+\ldots \ldots+x^{22}=\left(1+x^{2}\right)\left(1+x^{4}\right)\left(1+x^{4}+x^{8}\right)\left(1-x^{4}+x^{8}\right) \\
Q(x)=1+x+x^{2}+x^{3}+\ldots \ldots+x^{11}=(1+x)\left(1+x^{2}\right)\left(1+x^{4}+x^{8}\right) \\
\frac{P(x)}{Q(x)}=\frac{\left(1+x^{4}\right)\left(1-x^{4}+x^{8}\right)}{(1+x)}=\frac{1-x^{4}+x^{8}+x^{4}-x^{8}+x^{12}}{1+x}=\frac{1+x^{12}}{1+x}
\end{array}
$$
Remainder. When $\left(1+x^{12}\right)$ is divided by $(1+x)$ is $=2$
Now remainder $\left.\mathrm{P}_{\Delta} \mathrm{x}\right)$ divided by $\mathrm{Q}(\mathrm{x})$
$$
\begin{array}{l}
=2\left(1+x^{2}\right)\left(1+x^{4}+x^{8}\right) \\
=2\left(1+x^{2}+\ldots \ldots+x^{10}\right)
\end{array}
$$

Topic: Mathematics, Quadratic Equation, KVPY, KVPY 2016 (SB/SX)
The remainder when the polynomial $2 x^5-3 x^4+5 x^3-3 x^2+7 x-9$ is divided by $x^2-x-3$ is

Options:
A $-41 x-3$
B $41 x+3$
C $41 x-3$
D $-41 x+3$

The correct answer is:
$41 x+3$

Solution:
$\begin{array}{r}\left.x^2-x-3\right) 2 x^5-3 x^4+5 x^3-3 x^2+7 x-9\left(2 x^3-x^2\right. \\ +10 x+4\end{array}$
$\therefore$ Remainder $=41 x+3$

Topic: Mathematics, Quadratic Equation, AP EAMCET, AP EAMCET 2022 (07 Jul Shift 1)
The renal tubule of the uriniferous tubule is mainly made up of ___________ epithelium.

Options:
A Squamous
B Cuboidal
C Columnar
D Ciliated

The correct answer is:
Cuboidal

Solution:
Epithelial tissue can be classified as simple or compound epithelium depending on number of layers resting on the basement membrane. The simple epithelium consists of only one layer resting on the basement membrane. This is further divided into various types depending on shape of cells. These are squamous epithelium when the cells are flat, cuboidal when the cells are square in shape, columnar when the cells are tall and pillar-like and ciliated when cells show presence of cilia. The main function of simple epithelium is transport of substances.

Topic: Biology, Excretory Products and their Elimination, NEET
The residue left after methane production from cattle dung is

Options:
A Burnt
B Buried in land fills
C Used as manure
D Used in civil construction.

The correct answer is:
Used as manure

Solution:
Biogas predominantly containing methane gas is produced by the action of microbes. Certain bacteria growing anaerobically produce methane gas in large quantities along with the release of CO2 and H2 and hence such bacteria are also known as methanogens. Excreta of cattle contain methanogens in large amounts and hence cattle dung is used in making biogas. Biogas is made in the biogas plant where cattle dung is added to the biowaste. Biogas is supplied to the nearby houses through the pipes while the remaining waste is used as manure. Manure is rich in fertilizer and hence used in agriculture.

Topic: Biology, Microbes in Human Welfare, NEET
The resistance of 0.01N solution of an electrolyte was found to be 220 ohm at 298âK using a conductivity cell with a cell constant of 0.88 cmâ1. The value of equivalent conductance of solution is-

Options:
A 400 mho cm2 g eqâ1
B 295Â mhoÂ cm2Â gÂ eqâ1
C 419Â mhoÂ cm2Â gÂ eqâ1
D 425Â mhoÂ cm2Â gÂ eqâ1

The correct answer is:
400 mho cm2 g eqâ1

Solution:
Λeq=k×100N=1R×la×1000N
=1R×cell constant×1000N
=1220×0.88×10000.01
=400 mho cm2g eq−1

Topic: Chemistry, Electrochemistry, JEE Main
The resistance of a bulb filament is $100 \Omega$ at a temperature of $100^{\circ} \mathrm{C}$. If its temperature coefficient of resistance be $0.005$ per ${ }^{\circ} \mathrm{C}$, its resistance will become $200 \Omega$ at a temperature of

Options:
A
$200^{\circ} \mathrm{C}$

B
$300^{\circ} \mathrm{C}$

C
$400^{\circ} \mathrm{C}$

D
$500^{\circ} \mathrm{C}$

The correct answer is:
$300^{\circ} \mathrm{C}$

Solution:
$200=100[1+(0.005 \times \Delta t)]$
$T-100=200$
$\mathrm{T}=300^{\circ} \mathrm{C}$

Topic: Physics, Current Electricity, JEE Main, JEE Main 2006
The resistance of a conductor at 15°C is 16 Ω and at 100°C is 20 Ω. What will be the temperature coefficient of resistance of the conductor?

Options:
A 0.010°C-1
B 0.033°C-1
C 0.003°C-1
D 0.042°C-1

The correct answer is:
0.003°C-1

Solution:
16=R01+α15-To

20=R01+α100-T0

Assuming To=0°C, as a general convention.

⇒1620=1+α×151+α×100

⇒α=0.003C∘-1

Topic: Physics, Current Electricity, JEE Main, JEE Main 2021 (27 Jul Shift 2)