datavorous/entrance-exam-dataset huggingface dataset:
The resistance of a galvanometer is 50 ohm and the maximum current which can be passed through it is 0.002 A. What resistance must be connected to it in order to convert it into an ammeter of range 0-0.5 A?

Options:
A 0.2 ohm
B 0.5 ohm
C 0.002 ohm
D 0.02 ohm

The correct answer is:
0.2 ohm

Solution:
RGig=Rsis
50×0.002=Rs0.498
Rs=0.2 Ohm

Topic: Physics, Current Electricity, JEE Main, JEE Main 2019 (09 Apr Shift 2)
The resistance of a galvanometer is $50 \mathrm{ohms}$ and the current required to give full scale deflection is $100 \mu \mathrm{A}$. In order to convert it into an ammeter, reading upte $10 \mathrm{~A}$, it is necessary to put a resistance of

Options:
A $5 \times 10^{-3} \Omega$ in parallel
B $5 \times 10^{-4} \Omega$ in parallel
C $10^5 \Omega$ in series
D $99,950 \Omega$ in series

The correct answer is:
$5 \times 10^{-4} \Omega$ in parallel

Solution:
Resistance in parallel $S=\frac{G i_g}{i-i_g}=\frac{50 \times 100 \times 10^{-6}}{\left(10-100 \times 10^{-6}\right)}$
$\Rightarrow S=5 \times 10^{-4} \Omega$

Topic: Physics, Current Electricity, JEE Main
The resistance of a wire at 20 °C is 20 Ω and 500 °C is 60 Ω. At which temperature, its resistance will be 25 Ω ?

Options:
A 50 °C
B 60 °C
C 70 °C
D 80 °C

The correct answer is:
80 °C

Solution:
We have, Rt=R0 1+αt

Here, 20=R0 1+20α

60=R0 1+500α

⇒ 6020=R0 1+500αR0 1+20α

3=1+500 α1+20 α

⇒ 3 1+20 α=1+500 α

⇒ 3+60α=1+500α

⇒ 440 α=2

⇒ α=2440=1220I°C

Also given Rt=25 °C, t= ?

Now, 2025=R0 1+20220R0 1+t220

4 1+t220=5 1+20220

On solving,

t=80 °C

Topic: Physics, Current Electricity, JEE Main
The resistance of a wire is 5 Î© at 50 Â°C and 6 Î© at 100 Â°C. The resistance of the wire at 0 Â°C will be

Options:
A 3 Ω
B 2 Ω
C 1 Ω
D 4 Ω

The correct answer is:
4 Ω

Solution:
Given: R50=5 Ω, R100=6 Ω
R t = R 0 1 + α t
Where Rt = resistance of a wire at t °C, R0 = resistance of a wire at 0 °C, α = temperature coefficient of resistance.
∴ R 5 0 = R 0 1 + α 5 0
and R 1 00 = R 0 1 + α 1 00
or R 5 0 - R 0 = R 0 α 5 0
R 1 00 - R 0 = R 0 α 1 00
Dividing (i) by (ii), we get
5 - R 0 6 - R 0 = 1 2
or 10-2R0=6-R0
or R0=4 Ω

Topic: Physics, Current Electricity, JEE Main
The resistance of a wire is 5 Ω. Its new resistance in ohm, if stretched to 5 times of its original length will be :

Options:
A 625
B 125
C 5
D 25

The correct answer is:
125

Solution:
The resistance of a wire (R) is related to its resistivity ρ, length(l) and cross-sectional area A as:

R=ρlA

As the length of wire increases, the cross-sectional area decreases. The cross-sectional area in terms of volume (V) and length (l) is

A=Vl

∴ R=ρl2V

Taking the volume of wire constant,

R∝l2

So if R' is the resistance of wire after getting redrawn and R is the initial resistance of wire,

∴ R'R=(5l)2l2=25⇒R'=25×R=125 Ω

Topic: Physics, Current Electricity, JEE Main, JEE Main 2023 (25 Jan Shift 2)
The resistance of a wire is $5 \mathrm{~ohm}$ at $50^{\circ} \mathrm{C}$ and $6 \mathrm{~ohm}$ at $100^{\circ} \mathrm{C}$. The resistance of the wire at $0^{\circ} \mathrm{C}$ will be

Options:
A
$2 \mathrm{~ohm}$

B
$1 \mathrm{~ohm}$

C
$4 \mathrm{~ohm}$

D
$3 \mathrm{~ohm}$

The correct answer is:
$4 \mathrm{~ohm}$

Solution:
$\frac{5}{6}=\frac{1+50 \alpha}{1+100 \alpha}$
$5=R_0(1+\alpha \times 50)$
$R_0=4$

Topic: Physics, Current Electricity, JEE Main, JEE Main 2007
The resistance of a wire is $10 \Omega$. Its length is increased by $10 \%$ by stretching. The new resistance will now be

Options:
A $12 \Omega$
B $1.2 \Omega$
C $13 \Omega$
D $11 \Omega$

The correct answer is:
$12 \Omega$

Solution:
Since $R \propto I^2 \Rightarrow$ If length is increased by $10 \%$, resistance is increases by almost $20 \%$
Hence new resistance $R=10+20 \%$ of 10 $=10+\frac{20}{100} \times 10=12 \Omega$

Topic: Physics, Current Electricity, JEE Main
The resistance of a wire is $20 \Omega$. It is stretched, so that the length becomes three times, then the new resistance of the wire will be

Options:
A $200 \Omega$
B $160 \Omega$
C $120 \Omega$
D $180 \Omega$

The correct answer is:
$180 \Omega$

Solution:
Initial resistance,
Now, $\quad l_2=3 l_1$
$\because$ Volume remains same.
$\begin{array}{ll}
\therefore \quad & V_1=V_2 \Rightarrow A_1 l_1=A_2 l_2 \\
& A_1 l_1=A_2 \times 3 l_1 \Rightarrow A_2=\frac{A_1}{3}
\end{array}$
So, final resistance,
$\begin{aligned} R_2=\frac{\rho l_2}{A_2} & =\frac{\rho \times 3 l_1}{\left(\frac{A_1}{3}\right)}=\frac{9 \rho l_1}{A_1}=9 \times 20 \Omega \quad \text { [by Eq. (i)] } \\ & =180 \Omega\end{aligned}$

Topic: Physics, Current Electricity, TS EAMCET, TS EAMCET 2020 (14 Sep Shift 1)
The resistance of a wire is $R$. It is bent at the middle by $180^{\circ}$ and both the ends are twisted together to make a shorter wire. The resistance of the new wire is

Options:
A $2 R$
B $R / 2$
C $R / 4$
D $R / 8$

The correct answer is:
$R / 4$

Solution:
Resistance of wire $(R)=\rho \frac{l}{A}$ If wire is bent in the middle then

$$

l^{\prime}=\frac{l}{2}, A^{\prime}=2 A

$$

$\therefore$ New resistance, $\mathrm{R}^{\prime}=\rho \frac{l^{\prime}}{\mathrm{A}^{\prime}}=\frac{\rho \frac{l}{2}}{2 A}=$

$\frac{\rho l}{4 A}=\frac{R}{4}$

Topic: Physics, Current Electricity, JEE Main
The resistance of a wire is $R$. It is bent at the middle by $180^{\circ}$ and both the ends are twisted together to make a shorter wire. The resistance of the new wire is

Options:
A
$2 R$

B
$R / 2$

C
$R / 4$

D
$R / 8$

The correct answer is:
$R / 4$

Solution:
Resistance of wire $(R)=\rho \frac{l}{A}$ If wire is bent in the middle then
$$
\begin{aligned}
& l^{\prime}=\frac{l}{2}, A^{\prime}=2 \mathrm{~A} \\
& \therefore \text { New resistance, } \mathrm{R}^{\prime}=\rho \frac{l^{\prime}}{\mathrm{A}^{\prime}} \\
& =\frac{\rho \frac{l}{2}}{2 A}=\frac{\rho l}{4 A}=\frac{R}{4} \text {. } \\
&
\end{aligned}
$$

Topic: Physics, Current Electricity, JEE Main, JEE Main 2012 (26 May Online)
The resistance of a wire is $R$ ohm. If it is melted and stretched to $n$ times its original length, its new resistance will be

Options:
A $n R$
B $\frac{R}{n}$
C $n^{2} R$
D $\frac{R}{n^{2}}$

The correct answer is:
$n^{2} R$

Solution:
Volume of material remain same after stretching. As volume remains same,
$$
\begin{aligned}
& & A_{1} l_{1} &=A_{2} l_{2} \\
& H e r e, \quad & l_{2} &=n l_{1} \\
\Rightarrow \quad & A_{2} &=\frac{A_{1} l_{1}}{l_{2}}=\frac{A_{1}}{n}
\end{aligned}
$$
Resistance of wire after stretching,
$$
R_{2}=\frac{\rho l_{2}}{A_{2}}=\rho \frac{n l_{1}}{A_{1} / n}=n^{2} \rho \frac{l_{1}}{R_{1}}=n^{2} R \quad\left(\because R=\rho \frac{l_{1}}{A_{1}}\right)
$$

Topic: Physics, Current Electricity, COMEDK, COMEDK 2021
The resistance of a wire is R ohm. If it is melted and stretched to n times its original length, its new resistance will be

Options:
A nR
B Rn
C n2R
D Rn2

The correct answer is:
n2R

Solution:
Volume of wire will remain constant
So, A1l1=A2l2
⇒A2=A1n
The resistance of a wire with a length l, an area of A, and resistivity ρ is stated as R=ρlA.

Now, the new length is nl.
∴ New resistance R'=ρnlA/n⇒R'=ρn2lA
⇒ R'=n2R

Topic: Physics, Current Electricity, NEET, NEET 2017
The resistance of an electrical toaster has a temperature dependence given by RT=R01+αT-T0 in its range of operation. At T0=300 K, R=100 Ω and at T=500 K, R=120 Ω. The toaster is connected to a voltage source at 200 V and its temperature is raised at a constant rate from 300 to 500 K in 30 s. The total work done in raising the temperature is :
Note: This question was awarded as the bonus since all options were incorrect in the exam.

Options:
A 60000 ln65 J
B 200 ln23 J
C 300 J
D 400 ln1.51.3 J

The correct answer is:
60000 ln65 J

Solution:
P= V 2 R =2002R0[1+α(T-T0)]
T → temperature at 't'
T 0 =300K → temperature at t = 0
T - T 0 = 5 0 0 - 3 0 0 3 0 t
T - T 0 = 2 0 0 3 0 t
T - T 0 = 2 0 t 3
α=1R0∆R∆T=110020200=10-3K
H=∫Pdt=
∫ 0 3 0 2 0 0 2 1 0 0 1 + α 2 0t 3 dt = 2 0 0 × 2 0 0 1 0 0 ∫ 0 3 0 dt 1 + 2 0 α 3 t
= 4 0 0 × 3 2 0 α ℓ n ( 1+ 20∝ 3 ×30 1 )
= 6 0 0 0 0 ℓ n 6 5

Topic: Physics, Current Electricity, JEE Main, JEE Main 2016 (10 Apr Online)
The resistance of an ideal voltmeter is

Options:
A Zero
B Very low
C Very large
D Infinite

The correct answer is:
Infinite

Solution:
The resistance of an ideal voltmeter is considered as infinite.

Topic: Physics, Current Electricity, JEE Main
The resistance of conductivity cell containing 0.001âMâKCl solution at 298âK is 1500 Î©. What is the cell constant if conductivity of 0.001âMâKCl solution at 298 K is 0.146Ã10-3 Scm-1.

Options:
A 0.200Â cm-1
B 0.219 cm-1
C 0.195Â cm-1
D 0.190Â cm-1

The correct answer is:
0.219 cm-1

Solution:
Cell constant =conductivityconductanâ¡ce
= Conductivity Ã Resistance
=0.146Ã10-3 Scm-1Ã1500 Î©
=0.219 cm-1

Topic: Chemistry, Electrochemistry, JEE Main
The resistance of conductivity cell with cell constant 1.14 cm-1, containing 0.001M KCl at 298 K is 1500Ω. The molar conductivity of 0.001M KCl solution at 298 K in S cm2 mol-1 is (Integer answer)

Options:

The correct answer is:
760

Solution:
k=1R×ℓA=11500×1.14 S cm-1

⇒∧m=1000×1.1415000.001 S cm2 mol-1

=760 S cm2 mol-1⇒760

Topic: Chemistry, Electrochemistry, JEE Main, JEE Main 2021 (27 Aug Shift 2)
The resistance of each straight section is r=2 Î©. Find the equivalent resistance (in ohms) between A and B.

Options:

The correct answer is:
7

Solution:
The circuit can be simplified as shown in the above figure. Due to symmetry, we can see that the distribution of current at point A in the circuit, is the same as the recombination of the current before point B. Also, the junction which is present between the points C and D can be disconnected because it won't affect the distribution of current.

rAB=r+1.5r+r

⇒rAB=3.5r=7 Ω

Topic: Physics, Current Electricity, JEE Main
The resistance of each straight section r=2 Î©. Find the equivalent resistance (in ohms) between A and B.

Options:

The correct answer is:
7

Solution:
⇒ rAB=r+1.5r+r=3.5r=7 Ω

Topic: Physics, Current Electricity, JEE Main
The resistance of hot tungsten filament is about 10 times the cold resistance. What will be the resistance of $100 \mathrm{~W}$ and $200 \mathrm{~V}$ lamp when not in use?

Options:
A
$40 \Omega$

B
$20 \Omega$

C
$400 \Omega$

D
$200 \Omega$

The correct answer is:
$40 \Omega$

Solution:
$R_{\text {hot }}=\frac{V^2}{P}=\frac{200 \times 200}{100}=400 \Omega$
cold resistance $=R_{\text {hot }} / 10=400 / 10=40 \Omega$

Topic: Physics, Current Electricity, JEE Main, JEE Main 2005
The resistance of the following circuit figure between A and B is

Options:
A 32 Ω
B 2 Ω
C 4 Ω
D 8Â Î©

The correct answer is:
32 Ω

Solution:
Since resistance connected in arms CE, ED, CF
and FD will form a balanced Wheatstone bridge,
therefore, the resistance of arm EF becomes
ineffective. Now resistance of arm CED or CFD=2+2=4Î©
Effective resistance of these two parallel arm =4Ã44+4=2Î©
Now resistance of arm ACDB=2+2+2=6Î©
in parallel with resistance arm AB=2Î© Thus,
effective resistance between A and B
=6Ã26+2=32Î©

Topic: Physics, Current Electricity, JEE Main
The resistance of the meter bridge AB in given figure is $4 \Omega$. With a cell of emf $\varepsilon=0.5 \mathrm{~V}$ and rheostat resistance $\mathrm{R}_{\mathrm{h}}=2 \Omega$ the null point is obtained at some point J. When the cell is replaced by another one of emf $\varepsilon=\varepsilon_{2}$ the same null point $\mathrm{J}$ is found for $\mathrm{R}_{\mathrm{h}}=6 \Omega$. The $\operatorname{emf} \varepsilon_{2}$ is:

Options:
A $0.4 \mathrm{~V}$
B $0.3 \mathrm{~V}$
C $0.6 \mathrm{~V}$
D $0.5 \mathrm{~V}$

The correct answer is:
$0.3 \mathrm{~V}$

Solution:
Given, Emf of cell, $\varepsilon=0.5 \mathrm{v}$ Rheostat resistance, $\mathrm{R}_{\mathrm{h}}=2 \Omega$ Potential gradient is

$\frac{d v}{d L}=\left(\frac{6}{2+4}\right) \times \frac{4}{L}$

Let null point be at $\ell \mathrm{cm}$ when cell of $\mathrm{emf} \varepsilon=0.5 \mathrm{v}$ is used.

thus $\varepsilon_{1}=0.5 \mathrm{~V}=\left(\frac{6}{2+4}\right) \times \frac{4}{\mathrm{~L}} \times \ell$...(i)

For resistance $\mathrm{R}_{\mathrm{h}}=6 \Omega$ new potential gradient is $\left(\frac{6}{4+6}\right) \times \frac{4}{L}$ and at null point

$\left(\frac{6}{4+6}\right)\left(\frac{4}{\mathrm{~L}}\right) \times \ell=\varepsilon_{2}$...(ii)

Dividing equation (i) by (ii) we get

$\frac{0.5}{\varepsilon_{2}}=\frac{10}{6}$ thus $\varepsilon_{2}=0.3 \mathrm{v}$

Topic: Physics, Experimental Physics, JEE Main, JEE Main 2019 (11 Jan Shift 1)
The resistance of the series combination of two resistances is $S$. When they are joined in parallel through total resistance is $P$. If $S=n P$, then the minimum possible value of $n$ is

Options:
A
4

B
3

C
2

D
1

The correct answer is:
4

Solution:
Let resistances be $R_1$ and $R_2$
So, $S=R_1+R_2$;
$$
\begin{aligned}
& \mathrm{P}=\frac{\mathrm{R}_1 \mathrm{R}_2}{\mathrm{R}_1+\mathrm{R}_2} \\
& \mathrm{~S}=\mathrm{nP} \\
& \mathrm{R}_1+\mathrm{R}_2=\frac{\mathrm{nR}_1 \mathrm{R}_2}{\mathrm{R}_1+\mathrm{R}_2} \\
& \left(\mathrm{R}_1+\mathrm{R}_2\right)^2=\mathrm{nR}_1 \mathrm{R}_2
\end{aligned}
$$
If $R_1=R_2$, so minimum value of $n=4$.

Topic: Physics, Current Electricity, JEE Main, JEE Main 2004
The resistance of the series combination of two resistors is S. When they are joined in parallel the equivalent resistance becomes P. If S=nP, then the minimum possible value of n is

Options:

The correct answer is:
4

Solution:
In series combination, S=R1+R2
In parallel combination, P = R 1 R 2 R 1 + R 2
∵ S=nP
∴ R 1 + R 2 = n R 1 R 2 R 1 + R 2
∴ R 1 + R 2 2 = n R 1 R 2
For minimum value, R1=R2=R
R+R2=nR×R⇒4R2=nR2⇒n=4

Topic: Physics, Current Electricity, JEE Main
The resistance of the wire is R1. It is now stretched (keeping the volume constant) without breaking so that its length becomes three times. Now the resistance becomes R2, then what is the value of R2: R1?

Options:

The correct answer is:
9

Solution:
For a given wire, its resistance

RâlA
â´ R1âl1r12

The wire is stretched such that l2=3l1

But density is unchanged.

â´Density =MV=M4Ïr12l1=M4Ïr22l2

âr22=l1r12l2=r123 â¦..(i) âµl2=3l1

Now, R2âl2r22

â´ R2â9l1r12 ....ii

From (i) and (ii)

â´ R2:R1=9:1

Topic: Physics, Current Electricity, JEE Main
The resistance per centimeter of a meter bridge wire is r, with X Ω resistance in left gap. Balancing length from left end is at 40 cm with 25 Ω resistance in right gap. Now the wire is replaced by another wire of 2r resistance per centimeter. The new balancing length for same settings will be at

Options:
A 20 cm
B 10 cm
C 80 cm
D 40 cm

The correct answer is:
40 cm

Solution:
For first case:

Xrl1=25rl2 ... (i)

For second case:

X2rl'1=252rl'2 ... (ii)

From (i) and (ii), we get

l'1l1=l'2l2 also total length is constant.

Therefore, balance length doesn't change and

l'1=l1=40 cm

Topic: Physics, Current Electricity, JEE Main, JEE Main 2024 (31 Jan Shift 2)
The resistance R=VI, where V=50±2 V and I=20±0.2 A. The percentage error in R is x %. The value of x to the nearest integer is ________.

Options:

The correct answer is:
5

Solution:
ΔRR×100=ΔVV×100+ΔII×100

% error in R=250×100+0.220×100

% error in R=4+1

% error in R=5%

Topic: Physics, Mathematics in Physics, JEE Main, JEE Main 2021 (16 Mar Shift 1)
The resistance R=VI, where V=(200±5) V and I=(20±0.2) A, the percentage error in the measurement of R is :

Options:
A 3.5%
B 7%
C 3%
D 5.5%

The correct answer is:
3.5%

Solution:
Given that, R=VI

According to error analysis

dRR=dVV+dII⇒dRR=5200+0.220⇒dRR=7200

Thus, the percentage error can be calculated as follows:

% error =dRR×100=7200×100=3.5%

Topic: Physics, Mathematics in Physics, JEE Main, JEE Main 2024 (29 Jan Shift 1)
The resistances of a wire at temperatures $t^{\circ} \mathrm{C}$ and $0^{\circ} \mathrm{C}$ are related by

Options:
A $R_{\mathrm{t}}=R_0(1+\alpha t)$
B $R_{\mathrm{t}}=R_0(1-\alpha t)$
C $R_{\mathrm{t}}=R_0^2(1+\alpha t)$
D $R_{\mathrm{t}}=R_0^2(1-\alpha t)$

The correct answer is:
$R_{\mathrm{t}}=R_0(1+\alpha t)$

Solution:
Correct option is A. $R_t=R_0(1+\alpha t)$
The resistance of a wire at temperatures $t^{\circ} C$ and $0^{\circ} C$ are related by
$R_t=R_0(1+\alpha t)$
where $R_o$ is temperature at $0^{\circ} C$ and $\alpha$ is a constant

Topic: Physics, Current Electricity, JEE Main
The resistances of the platinum wire of a platinum resistance thermometer at the ice point and steam point are $8 \Omega$ and $10 \Omega$ respectively. After inserting in a hot bath of temperature $400^{\circ} \mathrm{C}$, the resistance of platinum wire is :

Options:
A $10 \Omega$
B $8 \Omega$
C $16 \Omega$
D $2 \Omega$

The correct answer is:
$16 \Omega$

Solution:
Given $R_0=8 \Omega, R_{100}=10 \Omega$
$\therefore \mathrm{R}_{100}=\mathrm{R}_0(1+\alpha \Delta \mathrm{T})$
Also, $R_{400}=R_0\left(1+\alpha \Delta \mathrm{T}^1\right)$
$\begin{aligned}
& \therefore 10=8(1+\alpha \times 100) \Rightarrow 100 \alpha=\frac{1}{4} \\
& \therefore \mathrm{R}_{400}=8(1+400 \alpha)=8(1+1)=16 \Omega
\end{aligned}$

Topic: Physics, Thermal Properties of Matter, JEE Main, JEE Main 2024 (04 Apr Shift 1)
The resistive network shown below is connected to a D.C. source of 16 V. The power consumed by the network is 4 Watt. The value of R is:

Options:
A 16 Ω
B 8 Ω
C 6 Ω
D 1 Ω

The correct answer is:
8 Ω

Solution:
Req for mixed series and parallel combination of

given circuit Req=114R+14R+R+116R+112R+R

Req=8R
Power dissipated by circuit P=V2Req
⇒4=1628R
⇒R=25632=8 Ω

Topic: Physics, Current Electricity, JEE Main, JEE Main 2019 (12 Apr Shift 1)
The resistivity of a 0.8M solution of an electrolyte is 5×10-3Ωcm. Its molar conductivity is 104Ω-1 cm2 mol-1. (Nearest integer)

Options:

The correct answer is:
25

Solution:
Given,
Resistivity (ρ)=5×10-3Ωcm

Molarity (M)=0.8M

Thus, Molar conductivity of electrolyte λM=1ρ×1000M

=15×10-3×10000.8

=0.25×106

=25×104 cm2 mol-1.

Topic: Chemistry, Electrochemistry, JEE Main, JEE Main 2023 (31 Jan Shift 2)
The resistivity (ρ) of semiconductor varies with temperature. Which of the following curve represents the correct behaviour?

Options:
A
B
C
D

The correct answer is:

Solution:
When a modest amount of energy is applied to semiconductors, electrons quickly travel to the conduction band and are then free to move across the crystal's lattice.

If thermal energy is provided, the conduction of electrons will increase. Hence, as the temperature is increased, the resistivity of the semiconductor is decreased according to the following diagram:

Topic: Physics, Semiconductors, JEE Main, JEE Main 2023 (06 Apr Shift 1)
The resonance frequency of LCR series a.c. circuit is f0. Now the capacitance is made 4 times, then the new resonance frequency will become

Options:
A f04
B 2f0
C f0
D f02

The correct answer is:
f02

Solution:
In LCR series circuit, resonance frequency f0 is given by
LÏ=1CÏâÏ2=1LC â´Ï=1LC=2Ïf0
â´f0=12ÏLC or f0â1C
When the capacitance of the circuit is made 4 times, its resonant frequency become f0â²
â´f0â²f0=C4C or f0â²=f02

Topic: Physics, Alternating Current, JEE Main
The respective displacements (in cm ) of two particles of a medium disturbed by a wave propagation at x 1 =0 and x 2 =1 cm are given by the equations: y 1 =2 sin 3πt
y 2 =2 sin (3πt-π/8)

The speed of the wave is

Options:
A 16 cm s-1
B 24 cm s-1
C 12 cm s-1
D 8 cm s-1

The correct answer is:
24 cm s-1

Solution:
Given Ï=3Ï

â´f=Ï2Ï=1.5

Also Îx=1.0 cm

Given Ï=2ÏÎ»ÎxâÏ8=2ÏÎ»Ã1

âÎ»=16 cmâv=fÎ»=16Ã1.5=24 cm sec-1

Topic: Physics, Waves and Sound, JEE Main
The respective number of significant figures for the numbers $23.023,0.0003$ and $2.1 \times 10^{-3}$ are

Options:
A
$5,1,2$

B
$5,1,5$

C
$5,5,2$

D
$4,4,2$

The correct answer is:
$5,1,2$

Solution:
No Solution Available

Topic: Physics, Mathematics in Physics, JEE Main, JEE Main 2010
the respective speeds of five molecules are $2,1.5,1.6,1.6$ and $1.2 \mathrm{~km} / \mathrm{sec}$. The most probable speed in $\mathrm{km} / \mathrm{sec}$ will be

Options:
A $2$
B $1.58$
C $1.6$
D $1.31$

The correct answer is:
$1.6$

Solution:
Since maximum number of molecules travel with speed $1.6 \mathrm{~km} / \mathrm{sec}$ so $v_{m p}=1.6 \mathrm{~km} / \mathrm{sec}$

Topic: Physics, Kinetic Theory of Gases, JEE Main
The respiratory quotient varies due to

Options:
A Respiratory substrate
B Light and O2
C Respiratory product
D Temperature

The correct answer is:
Respiratory substrate

Solution:
The value of R.Q. always depends upon the chemical nature of the substrate. In respiration, many types of high energy compounds are oxidized. These are called respiratory substrate or respiratory fuel and include carbohydrates, fats and protein.
Carbohydrates such as glucose, fructose (hexoses), sucrose (disaccharide) or starch, insulin, hemicellulose (polysaccharide), etc; are the main substrates. The energy present in one gram carbohydrate is 4.4 Kcal or 18.4 kJ. Under certain conditions (mainly when carbohydrate reserves have been exhausted) fats are also oxidized. The energy present in one gram of fats is 9.8 Kcal or 41kJ. In the absence of carbohydrates and fats, protein also serves as respiratory substrate. The energy present in one gram of protein is 4.8 Kcal or 20 kJ.

Topic: Biology, Respiration in Plants, NEET
The restoring force of a spring with a block attached to the free end of the spring is represented by:

Options:
A

B

C

D

The correct answer is:

Solution:
Restoring force of spring,
$\mathrm{F} \propto-x$
(where, $x$ is the displacement)
Only graph (4) is representing the above equation.

Topic: Physics, Laws of Motion, JEE Main
The resultant force (in Î¼N) on the current loop PQRS due to a long current-carrying conductor (current = 20 A) will be

20 A â

Options:

The correct answer is:
500

Solution:
Force on SR and PQ are equal but opposite so their net will be zero.
Force between two parallel conductors carrying currents I1 and I2
F=μ02πI1I2lr
Where r= distance between two parallel conductors
FPS=10-7×2×20×20×15×10-22×10-2
=6×10-4 N
FQR=10-7×2×20×20×15×10-212×10-2
=1×10-4 N
Fnet=FPS-FQR
=6×10-4-1×10-4=5×10-4 N=500 μN

Topic: Physics, Magnetic Effects of Current, JEE Main
The resultant force of $5 \mathrm{~N}$ and $10 \mathrm{~N}$ can not be

Options:
A $12 \mathrm{~N}$
B $8 \mathrm{~N}$
C $4 \mathrm{~N}$
D $5 \mathrm{~N}$

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

Solution:
$F_{\max }=5+10=15 \mathrm{~N}$ and $F_{\min }=10-5=5 \mathrm{~N}$ Range of resultant $5 \leq F \leq 15$

Topic: Physics, Laws of Motion, JEE Main
The resultant force on the current loop PQRS due to a long current-carrying conductor(current = 20 A) will be

20 A â

Options:
A 10-4Â N
B 3.6Ã10-4Â N
C 1.8Ã10-4Â N
D 5Ã10-4 N

The correct answer is:
5Ã10-4 N

Solution:
Force on SR and PQ are equal but opposite so their net will be zero.
Force between two parallel conductors carrying currents I1 and I2
F=Î¼02ÏI1I2lr
Where r= distance between two parallel conductors
FPS=10-7Ã2Ã20Ã20Ã15Ã10-22Ã10-2
=6Ã10-4 N
FQR=10-7Ã2Ã20Ã20Ã15Ã10-212Ã10-2
=1Ã10-4 N
Fnet=FPS-FQR
=6Ã10-4-1Ã10-4=5Ã10-4 N

Topic: Physics, Magnetic Effects of Current, JEE Main
The resultant gate and its Boolean expression for the given circuit is

Options:
A OR, $A+B$
B $\mathrm{NOR}, \overline{A+B}$
C $\quad \mathrm{NAND}, \overline{A \cdot B}$
D $\quad A N D, A \cdot B$

The correct answer is:
OR, $A+B$

Solution:
Here a NOR gate is followed by a NOT gate which gives an OR gate which has Boolean expression $A+B$

Topic: Physics, Semiconductors, JEE Main
The resultant of $\vec{A}+\vec{B}$ is $\vec{R}_1$. On reversing the vector $\vec{B}$, the resultant becomes $\vec{R}_2$. What is the value of $R_1^2+R_2^2$

Options:
A $A^2+B^2$
B $A^2-B^2$
C $2\left(A^2+B^2\right)$
D $2\left(A^2-B^2\right)$

The correct answer is:
$2\left(A^2+B^2\right)$

Solution:
$\overrightarrow{\mathrm{R}}_1$=$\overrightarrow{\mathrm{A}}+\overrightarrow{\mathrm{B}}$, $\overrightarrow{\mathrm{R}}_2$= $\overrightarrow{\mathrm{A}}-\overrightarrow{\mathrm{B}}$
$R_1^2+R_2^2=\left(\sqrt{A^2+B^2}\right)^2+\left(\sqrt{A^2+B^2}\right)^2=2\left(A^2+B^2\right)$

Topic: Physics, Mathematics in Physics, JEE Main