datavorous/entrance-exam-dataset huggingface dataset:
Two radioactive substances X and Y originally have N1 and N2 nuclei respectively. Half life of X is half of the half life of Y. After three half lives of Y, number of nuclei of both are equal. The ratio N1N2 will be equal to :

Options:
A 13
B 81
C 31
D 18

The correct answer is:
81

Solution:
Tx=t;Ty=2t

3Ty=6t,

N1'=N2'

N1e-λ16t=N2e-λ26t

N1 N2=eλ1-λ26t=e1t-12t×6tln2=eln23=eln8=8

Topic: Physics, Nuclear Physics, JEE Main, JEE Main 2021 (25 Feb Shift 1)
Two rain drops reach the earth with different terminal velocities having ratio 9 : 4. Then the ratio of their volumes is

Options:
A 3 : 2
B 4 : 9
C 9 : 4
D 27 : 8

The correct answer is:
27 : 8

Solution:
We know that terminal velocity is
given by
VT=29nr2ρs-ρl
∴VT∝r⁡2
∴94=r⁡1r⁡22
∴ r⁡ 1 r⁡ 2 = 3 2
∴ V 1 V⁡ 2 = r⁡ 1 r⁡ 2 3 = 3 2 3
V1V2=27:8

Topic: Physics, Mechanical Properties of Fluids, JEE Main
Two reactions A1 and A2 have identical pre-exponential factors. The activation energy of A1 is more than A2 by 10 kJ mol-1 . If k1 and k2 are the rate constants for reactions A1 and A2, respectively at 300 K, then ln⁡k2k1 is equal to
R=8.314 J mol-1K-1

Options:
A 12
B 6
C 4
D 8

The correct answer is:
4

Solution:
From Arrhenius equation
k=Ae-Ea/RT
k1=Ae-Ea1/RT
k2=Ae-Ea2/RT
lnk2k1=eEa1-Ea2/RT
ln⁡k2k1=Ea1-Ea2RT=10×10008.314×300=4

Topic: Chemistry, Chemical Kinetics, JEE Main, JEE Main 2017 (02 Apr)
Two reactions are given below:
2Fe(s)+32O2( g)→Fe2O3( s),ΔHo=-822 kJ/mol

C(s)+12O2( g)→CO(g),ΔHo=-110 kJ/mol

Then enthalpy change for following reaction
3C(s)+Fe2O3( s)→2Fe(s)+3CO(g)

Options:

The correct answer is:
492

Solution:
Enthalpy change is the standard enthalpy of formation, in any general chemical reaction, the reactants undergo chemical changes and combine to give products. It can be represented by the following equation:

Reactant→Product

From the given equations,

2Fe(s)+32O2( g)→Fe2O3( s),ΔHo=-822 kJ/mol .........(i)

C(s)+12O2( g)→CO(g),ΔH°=-110 kJ/mol....(ii)

3C(s)+Fe2O3( s)→2Fe(s)+3CO(g),ΔH3=?

The value of ΔH3 can be calculated as,

ΔH3=3×ΔH2-ΔH1

=3(-110)+822

=492 kJ/mole

Topic: Chemistry, Thermodynamics (C), JEE Main, JEE Main 2024 (30 Jan Shift 2)
Two rectangular blocks A and B of different metals have the same length and the same area of cross-section. They are kept with their lengths aligned and their cross-sections in contact. The temperature at one end of A is 100 Â°C and that of at the other end B is 0 Â°C. If the ratio of their thermal conductivity is 1:3, then under steady-state, what will be the temperature (in Â°C) of the junction?

Options:

The correct answer is:
25

Solution:
it is given that K1K2=13⇒K1=K then K2=3K
The temperature of the junction in contact
θ=K1θ1+K2θ2K1+K2=1×100+3×01+3=1004=25 °C

Topic: Physics, Thermal Properties of Matter, JEE Main
Two resistance of 100Ω and 200Ω are connected in series with a battery of 4 V and negligible internal resistance. A voltmeter is used to measure voltage across 100Ω resistance, which gives reading as 1 V. The resistance of voltmeter must be _______ Ω.

Options:

The correct answer is:
200

Solution:
Voltage across 200 Ω=4-1=3 V.

Therefore, current through the battery, 3200

Now, equivalent resistance of RV & 100 Ω,=Rv100Rv+100

For voltmeter, Rv100Rv+100×3200=1

⇒3Rv=2Rv+200

⇒Rv=200 Ω

Topic: Physics, Current Electricity, JEE Main, JEE Main 2024 (30 Jan Shift 2)
Two resistances are given as $R_1=(10 \pm$ $0.5) \Omega$ and $R_2=(15 \pm 0.5) \Omega$. The percentage error in the measurement of equivalent resistance when they are connected in parallel is

Options:
A
6.33
B
2.33
C
4.33
D
5.33

The correct answer is:
4.33

Solution:
Equivalent resistance in parallel combination is given by
$\begin{aligned}
& \frac{1}{R_{\text {eq }}}=\frac{1}{R_1}+\frac{1}{R_2} \\
& \Rightarrow R_{\text {eq }}=\frac{R_1 R_2}{R_1+R_2}=\frac{10 \times 15}{10+15}=6
\end{aligned}$
Differentiating both sides, we get
$\begin{aligned} & \frac{\Delta R_{\text {eq }}}{\mathrm{R}_{\mathrm{eq}}^2}=\frac{\Delta \mathrm{R}_1}{\mathrm{R}_1^2}+\frac{\Delta \mathrm{R}_2}{\mathrm{R}_2^2} \\ & \Rightarrow \frac{\Delta \mathrm{R}_{\mathrm{eq}}}{\mathrm{R}_{\mathrm{eq}}}=\left(\frac{\Delta \mathrm{R}_1}{\mathrm{R}_1^2}+\frac{\Delta \mathrm{R}_2}{\mathrm{R}_2^2}\right) \mathrm{R}_{\mathrm{eq}} \\ & =\left(\frac{0.5}{100}+\frac{0.5}{225}\right) \cdot 6=\left(\frac{6 \times 0.5}{25}\right)\left(\frac{1}{4}+\frac{1}{9}\right)=\frac{13}{300} \\ & \frac{\Delta \mathrm{R}_{\mathrm{eq}}}{\mathrm{R}_{\mathrm{eq}}} \times 100=\frac{13}{3}=4.33 \%\end{aligned}$

Topic: Physics, Mathematics in Physics, JEE Main
Two resistances are measured in ohm and are given as:
R1=3 Î©Â±1%
R2=6 Î©Â±2%
When they are connected in parallel, the percentage error in equivalent resistance is

Options:
A 3%
B 4.5%
C 0.67%
D 1.33%

The correct answer is:
1.33%

Solution:
1Req=1R1+1R2
dReqR2=dR1R12+dR2R22⇒dReqReq=dR1R12+dR2R22×Req
⇒13+26×2
⇒∴ % error =43%

Topic: Physics, Units and Dimensions, JEE Main
Two resistors 400 Ω and 800 Ω are connected in series across a 6∨ battery. The potential difference measured by a voltmeter of 10 kΩ across 400 Ω resistor is close to:

Options:
A 2 V
B 1.8 V
C 2.05 V
D 1.95 V

The correct answer is:
1.95 V

Solution:
Let voltmeter reading is v

v100×400+v10000+v400 800=6

⇒ v+8v100+2v=6 ; 77v25=6 ; v=15077=1.95 v

Topic: Physics, Current Electricity, JEE Main, JEE Main 2020 (03 Sep Shift 2)
Two resistors are connected in series across a battery as shown in figure. If a voltmeter of resistance 2000 Ω is used to measure the potential difference across 500 Ω resister, the reading of the voltmeter will be _____ V

Options:

The correct answer is:
8

Solution:
The voltmeter resistance is given to be 2000 Ω. The equivalent resistance in the given circuit is Req=500×20002000+500+600=1000 Ω.

Now, current in the circuit is i=201000 A.

Reading of the voltmeter will be, V=20-600×201000=8 V

Topic: Physics, Current Electricity, JEE Main, JEE Main 2022 (29 Jun Shift 2)
Two resistors are connected in series in parallel. The equivalent resistance in the two cases are 9 ohm and 2 ohm respectively. Then the resistances of the component resistors are

Options:
A $2 \mathrm{ohm}$ and $7 \mathrm{ohm}$
B $3 \mathrm{ohm}$ and $6 \mathrm{ohm}$
C $3 \mathrm{ohm}$ and $9 \mathrm{ohm}$
D $5 \mathrm{ohm}$ and $4 \mathrm{ohm}$

The correct answer is:
$3 \mathrm{ohm}$ and $6 \mathrm{ohm}$

Solution:
$\begin{aligned} & R_1+R_2=9 \text { and } \frac{R_1 R_2}{R_1+R_2}=2 \Rightarrow R_1 R_2=18 \\ & R_1-R_2=\sqrt{\left(R_1+R_2\right)^2-4 R_1 R_2}=\sqrt{81-72}=3 \\ & R_1=6 \Omega, R_2=3 \Omega\end{aligned}$

Topic: Physics, Current Electricity, JEE Main
Two resistors of resistances \( R_{1}=(300 \pm 3) \Omega \) and \( R_{2}=(500 \pm 4) \Omega \) are connected in series. The equivalent resistance of the series combination is

Options:
A \( (800 \pm 1) \Omega \)
B \( (800 \pm 7) \Omega \)
C \( (200 \pm 7) \Omega \)
D \( (200 \pm 1) \Omega \)

The correct answer is:
\( (800 \pm 7) \Omega \)

Solution:
The equivalent resistance of series combination is
Rs=R1+R2=300 Ω+500 Ω=800 Ω
The error in equivalent resistance is given by
∆R=±∆R1+∆R2=±3+4 Ω=±7 Ω
Hence, the equivalent resistance along with error is (800±7) Ω

Topic: Physics, Current Electricity, JEE Main
Two resistors R1=(4±0.8) Ω and R2=(4±0.4) Ω are connected in parallel. The equivalent resistance of their parallel combination will be :

Options:
A (4±0.4) Ω
B (2±0.4) Ω
C (4±0.3) Ω
D (2±0.3) Ω

The correct answer is:
(2±0.3) Ω

Solution:
ΔRR2=ΔR1R12+ΔR2R22
ΔR=R2ΔR1R12+ΔR2R22
=220.842+0.442
=[0.2+0.1]=0.3
Rnet =R+ΔR=2±0.3

Topic: Physics, Current Electricity, JEE Main, JEE Main 2021 (01 Sep Shift 2)
Two rigid boxes containing different ideal gases are placed on a table. Box A contains one mole of nitrogen at temperature $T_0$, while Box $B$ contains one mole of helium at temperature $(7 / 3) T_0$. The boxes are then put into thermal contact with each other and heat flows between them until the gases reach a common final temperature. (Ignore the heat capacity of boxes). Then, the final temperature of the gases, $T_f$, in terms of $T_0$ is

Options:
A
$\mathrm{T}_{\mathrm{f}}=\frac{5}{2} \mathrm{~T}_0$

B
$\mathrm{T}_{\mathrm{f}}=\frac{3}{7} \mathrm{~T}_0$

C
$\mathrm{T}_{\mathrm{f}}=\frac{7}{3} \mathrm{~T}_0$

D
$T_f=\frac{3}{2} T_0$

The correct answer is:
$T_f=\frac{3}{2} T_0$

Solution:
$\Delta \mathrm{U}=0$
$\Rightarrow \frac{3}{2} \mathrm{R}\left(\mathrm{T}_{\mathrm{f}}-\mathrm{T}_0\right)+1 \times \frac{5}{2} \mathrm{R}\left(\mathrm{T}_{\mathrm{f}}-\frac{7}{3} \mathrm{~T}_0\right)=0$
$\mathrm{~T}_{\mathrm{f}}=\frac{3}{2} \mathrm{~T}_0$

Topic: Physics, Kinetic Theory of Gases, JEE Main, JEE Main 2006
Two rigid boxes containing different ideal gases are placed on a table. Box A contains one mole of nitrogen at temperature T0, while box B contains one mole of helium at temperature 73T0. The boxes are then put into thermal contact with each other, and heat flows between them until the gases reach a common final temperature (Ignore the heat capacity of boxes). Then , the final temperature of the gases, Tf, in terms of T0 is

Options:
A Tf=37T0
B Tf=73T0
C Tf=32T0
D Tf=52T0

The correct answer is:
Tf=32T0

Solution:
Here, change in internal energy of the system is zero, ie, increase in internal energy of one is equal to decrease in internal energy of other.
âUA=1×5R2(Tf-To)
âUB=1×3R2(Tf-73T0)
Now, âUA+âUB=0
5R2Tf-T0+3R2Tf-7T03=0
5Tf-5T0+3Tf-7T0=0
⇒ 8Tf=12T0
⇒ Tf=128T0=32T0

Topic: Physics, Thermodynamics, JEE Main
Two rods $A$ and $B$ are of equal lengths. Their ends are kept between the same temperature and their area of cross-sections are $A_1$ and $A_2$ and thermal conductivities $K_1$ and $K_2$. The rate of heat transmission in the two rods will be equal, if

Options:
A $K_1 A_2=K_2 A_1$
B $K_1 A_1=K_2 A_2$
C $K_1=K_2$
D $K_1 A_1^2=K_2 A_2^2$

The correct answer is:
$K_1 A_1=K_2 A_2$

Solution:
$\left(\frac{Q}{t}\right)_1=\frac{K_1 A_1\left(\theta_1-\theta_2\right)}{I}$ and $\left(\frac{Q}{t}\right)_2=\frac{K_2 A_2\left(\theta_1-\theta_2\right)}{1}$
given $\left(\frac{Q}{t}\right)_1=\left(\frac{Q}{t}\right)_2 \Rightarrow K_1 A_1=K_2 A_2$

Topic: Physics, Thermal Properties of Matter, JEE Main
Two rods A and B of identical dimensions are at a temperature of 30 Â°C. If A is heated upto 180 Â°C and B up to T (measured in Â°C) then the new lengths are the same. If the ratio of the coefficients of linear expansion of A and B is 4:3, then the value of T is

Options:
A 230 °C
B 200 °C
C 270 °C
D 250 °C

The correct answer is:
230 °C

Solution:
∆L1=∆L2
Lα1×150=Lα2×∆T2
⇒ ∆T2=α1α2×150=43×150=200 °C
T=200 °C+30 °C=230 °C

Topic: Physics, Thermal Properties of Matter, JEE Main
Two rods $A$ and $B$ of identical dimensions are at temperature $30^{\circ} \mathrm{C}$. If $\mathrm{A}$ is heated upto $180^{\circ} \mathrm{C}$ and $\mathrm{B}$ upto $\mathrm{T}^{\circ} \mathrm{C},$ then the new lengths are the same. If the ratio of the coefficients of linear expansion of $\mathrm{A}$ and $\mathrm{B}$ is $4: 3,$ then the value of $T$ is

Options:
A $230^{\circ} \mathrm{C}$
B $270^{\circ} \mathrm{C}$
C $200^{\circ} \mathrm{C}$
D $250^{\circ} \mathrm{C}$

The correct answer is:
$230^{\circ} \mathrm{C}$

Solution:
Change in length in both rods are same i.e.

$\Delta \ell_{1}=\Delta \ell_{2}$

$\ell \alpha_{1} \Delta \theta_{1}=\ell \alpha_{2} \Delta \theta_{2}$

$\begin{array}{l}

\frac{\alpha_{1}}{\alpha_{2}}=\frac{\Delta \theta_{2}}{\Delta \theta_{1}} \quad\left[\because \frac{\alpha_{1}}{\alpha_{2}}=\frac{4}{3}\right] \\

\frac{4}{3}=\frac{\theta-30}{180-30}

\end{array}$

$\theta=230^{\circ} \mathrm{C}$

Topic: Physics, Thermal Properties of Matter, JEE Main, JEE Main 2019 (11 Jan Shift 2)
Two rods of different materials having coefficients of linear expansion $\alpha_1, \alpha_2$ and Young's moduli $Y_1$ and $Y_2$ respectively are fixed between two rigid massive walls. The rods are heated such that they undergo the same increase in temperature. There is no bending of rods. If $\alpha_1: \alpha_2=2: 3$, the thermal stresses developed in the two rods are equally provided $Y_1: Y_2$ is equal to

Options:
A $2: 3$
B $1: 1$
C $3: 2$
D $4: 9$

The correct answer is:
$3: 2$

Solution:
Thermal stress $=Y \alpha \Delta \theta$.
If thermal stress and rise in temperature are equal then $Y\propto \frac{1}{\alpha} \Rightarrow \frac{I_1}{I_2}=\frac{\alpha_2}{\alpha_1}=\frac{3}{2}$

Topic: Physics, Mechanical Properties of Solids, JEE Main
Two rods of length \( a \) and \( b \) slide along the axes in a manner that their ends are always concyclic. If the locus of the centre of the circle passing through these ends is \( k\left(x^{2}-y^{2}\right)=a^{2}-b^{2} \), then find the value of \( k \).

Options:

The correct answer is:
4

Solution:
Let AB and CD be two rods of length a and b sliding along x- axis and y-axis respectively.

Let the equation of circle through A,B,C and D.

x2+y2+2gx+2fy+c=0

AB=x- intercept and circle

a=2g2-c i.e. a2=4g2-c ...1

CD=y- intercept and circle

b=2f2-c i.e. b2=4f2-c ...2

From 1 and 2

a2-b2=4g2-f2

Locus of centre is 4x2-y2=a2-b2

k=4

Topic: Mathematics, Circle, JEE Main
Two rods of the same length and material transfer a given amount of heat in 12 s, when they are joined end to end, i.e. in series. When they are joined in parallel, they will transfer the same heat under the same conditions in

Options:
A 24 s
B 3 s
C 48 s
D 1.5 s

The correct answer is:
3 s

Solution:
∆t=∆Q∆xKA∆T
When two rods of same length are joined in parallel,
A →2 and ∆x→12 times
∴ ∆t becomes14 times i.e.,14×12 s=3 s

Topic: Physics, Thermal Properties of Matter, JEE Main
Two rods one made of copper and other made of steel of the same length and same cross sectional area are joined together. The thermal conductivity of copper and steel are $385 \mathrm{~J} \mathrm{~s}^{-1} \mathrm{~K}^{-1} \mathrm{~m}^{-1}$ and $50 \mathrm{~J} \mathrm{~s}^{-1} \mathrm{~m}^{-1}$ respectively. The free ends of copper and steel are held at $100^{\circ} \mathrm{C}$ and $0^{\circ} \mathrm{C}$ respectively. The temperature at the junction is, nearly:

Options:
A $12^{\circ} \mathrm{C}$
B $50^{\circ} \mathrm{C}$
C $73^{\circ} \mathrm{C}$
D $88.5^{\circ} \mathrm{C}$

The correct answer is:
$88.5^{\circ} \mathrm{C}$

Solution:
As per question, we have;
$\begin{aligned}
& \theta=\text { ? } \\
& \text {As } \\
& \mathrm{K}=\frac{Q d}{\mathrm{~A} \Delta t}, \\
& \text {where, } \mathrm{K}=\text { thermal conductivity } \\
& Q=\text { amount of heat transferred } \\
& d=\text { distance between two planes } \\
& \mathrm{A}=\text { area of surface, and } \\
& \Delta \mathrm{T}=\text { difference in Temp. } \\
& \frac{\Delta \mathrm{Q}}{\Delta t}=\frac{385 \times \mathrm{A} \times\left(100^{\circ} \mathrm{C}-\theta\right)}{l} \\
& =\frac{50 \times \mathrm{A} \times\left(\theta-0^{\circ} \mathrm{C}\right)}{l} \\
& \Rightarrow 385\left(100^{\circ} \mathrm{C}-\theta\right)=50\left(\theta-0^{\circ} \mathrm{C}\right) \\
& \Rightarrow \theta=88.5^{\circ} \mathrm{C}
\end{aligned}$

Topic: Physics, Thermal Properties of Matter, JEE Main
Two rods, one of aluminium and the other made of steel, having initial lengths l1 and l2 are connected together to form a single rod of length l1+l2. The coefficients of linear expansion for aluminium and steel are Î±a and Î±s respectively. If the length of each rod increases by the same amount when their temperature is raised by t Â°C, then find the ratio l1l1+l2

Options:
A α s α a
B α a α s
C Î± s ( Î± a + Î± s )
D α a ( α a + α s )

The correct answer is:
α s ( α a + α s )

Solution:
Given, Δl1=Δl2
or, l1αat=l2αst
∴l1l2=αsαa
or, l1l1+l2=αsαa+αs

Topic: Physics, Thermal Properties of Matter, JEE Main
Two rods (one semi-circular and other straight) of the same material and of same cross-sectional area are joined as shown in the diagram. The points A and B are maintained at different temperatures. The ratio of the heat transferred through a cross-section of a semi-circular rod to the heat transferred through a cross-section of the straight rod in a given time is

Options:
A 2 : π
B 1 : 2

C π : 2
D 3 : 2

The correct answer is:
2 : π

Solution:
dQ dt = KA Δ θ lâ¡
For both rods K, A and Δ θ are same.
∴ dQ dt ∝ 1 lâ¡
So, dQ / dt semicircular dQ / dt straight = lâ¡ straight lâ¡ semicircular = 2 r π r = 2 π

Topic: Physics, Thermal Properties of Matter, JEE Main
Two roots of the cubic \( x^{3}+3 x^{2}+k x-12=0 \) are real and unequal but have the same absolute value. Then the value of \( \mathrm{k} \) is

Options:
A \( 4 \)
B \( -4 \)
C \( 6 \)
D \( -9 \)

The correct answer is:
\( -4 \)

Solution:
Let the roots are r,-r and t

Hence, r-r+t=-3⇒t=-3

Now, product of the roots is (r)(-r)(t)=12

⇒-r2(-3)=12

⇒r2=4

Hence, roots are 2,-2 and -3

Now, sum of the product taken two at a time is
r(-r)+(-r)t+tr=k

⇒-4-6+6=k

⇒k=-4

Topic: Mathematics, Quadratic Equation, JEE Main
Two rotating bodies $\mathrm{A}$ and $\mathrm{B}$ of masses $\mathrm{m}$ and $2 \mathrm{~m}$ with moments of inertia $\mathrm{I}_{\mathrm{A}}$ and $\mathrm{I}_B\left(\mathrm{I}_{\mathrm{B}}\gt\mathrm{I}_{\mathrm{A}}\right)$ have equal kinetic energy of rotation. If $\mathrm{L}_{\mathrm{A}}$ and $\mathrm{L}_B$ be their angular momenta respectively, then

Options:
A $\mathrm{L}_{\mathrm{A}}\gt\mathrm{L}_{\mathrm{B}}$
B $\mathrm{L}_{\mathrm{A}}=2 \mathrm{~L}_{\mathrm{B}}$
C $\mathrm{L}_{\mathrm{B}}\gt\mathrm{L}_{\mathrm{A}}$
D $\mathrm{L}_{\mathrm{A}}=\frac{\mathrm{L}_{\mathrm{B}}}{2}$

The correct answer is:
$\mathrm{L}_{\mathrm{B}}\gt\mathrm{L}_{\mathrm{A}}$

Solution:
The kinetic energy of rotation is given by $\frac{1}{2} \mathrm{I} \omega^2$.
Since the kinetic energy of rotation is same,
$\begin{aligned}& \mathrm{I}_{\mathrm{A}} \omega_{\mathrm{A}}^2=\mathrm{I}_{\mathrm{B}} \omega_{\mathrm{B}}^2 \\& \Rightarrow \frac{\omega_{\mathrm{A}}}{\omega_{\mathrm{B}}}=\sqrt{\frac{\mathrm{I}_{\mathrm{B}}}{\mathrm{I}_{\mathrm{A}}}}\end{aligned}$
The ratio of the angular momentum becomes,
$\begin{aligned}& \frac{\mathrm{L}_{\mathrm{A}}}{\mathrm{L}_{\mathrm{B}}}=\frac{\mathrm{I}_{\mathrm{A}} \omega_{\mathrm{A}}}{\mathrm{I}_{\mathrm{B}} \omega_{\mathrm{B}}} \\& =\sqrt{\frac{\mathrm{I}_{\mathrm{A}}}{\mathrm{I}_{\mathrm{B}}}} \lt 1 \\& \Rightarrow \mathrm{L}_{\mathrm{B}}\gt\mathrm{L}_{\mathrm{A}}\end{aligned}$

Topic: Physics, Rotational Motion, JEE Main
Two salts A2X and MX have the same value of solubility product of 4.0×10-12. The ratio of their molar solubilizes i.e. SA2XS(MX)= ________.

(Round off to the Nearest Integer).

Options:

The correct answer is:
50

Solution:
For A2X

A2X→2A++X2-2S1S1Ksp=4 S13=4×10-12

S1=10-4

for MX

MX→M++X-S2S2

Ksp=S22=4×10-12

S2=2×10-6

So SA2xSMX=10-42×10-6=50

Topic: Chemistry, Ionic Equilibrium, JEE Main, JEE Main 2021 (16 Mar Shift 1)
Two satellite $A$ and $B$ go round a planet in circular orbits having radii $4 R$ and $R$ respectively. If the speed of $\mathrm{A}$ is $3 v$, the speed of $\mathrm{B}$ will be :

Options:
A $3 v$
B $6 v$
C $\frac{4}{3} v$
D $12 v$

The correct answer is:
$6 v$

Solution:
$\begin{aligned} & \mathbf{v}=\sqrt{\frac{G M}{R}} \\ & \frac{v_A}{v_B}=\sqrt{\frac{R_B}{R_A}}=\sqrt{\frac{R}{4 R}}=\frac{1}{2} \\ & v_B=2 v_A=6 v\end{aligned}$

Topic: Physics, Gravitation, JEE Main, JEE Main 2024 (08 Apr Shift 2)
Two satellites A and B are revolving around a planet in coplanar and concentric circular orbits of radii R1 and R2 respectively in the same sense. Their respective periods of revolution are 1 hour and 8 hours respectively. The radius of the orbit of A is equal to 104 km. What is their relative speed when they are closest?

Options:
A π×104 km h-1
B π2×104 km h-1
C 3π2×104 km h-1
D 2π×104 km h-1

The correct answer is:
π×104 km h-1

Solution:
By Kepler's third law T2∝R3 or T2R3=constant
∴ T12T22=R13R23
∴ 164=R13R23
∴ R1R2=14
∴ R2=4R1=4×104 km
Distance travelled by A in one revolution is S1=2πR1=2π×104 km
∴v1=S1T1=2π×1041=2π×104 km h-1
∴ Similarly, v2=S2T2=2π×4×1048=π×104 km h-1
V12=V1−V2=π×104 km h-1

Topic: Physics, Gravitation, JEE Main
Two satellites A and B having masses in the ratio 4:3 are revolving in circular orbits of radii 3r and 4r respectively around the earth. The ratio of total mechanical energy of A to B is

Options:
A 9:16
B 16:9
C 1:1
D 4:3

The correct answer is:
16:9

Solution:
Given that m1 m2=43,r1r2=34

Now the total energy of a satellite TE=12mvo2+-GMmr

Since the orbital speed of a satellite vo=GMr

⇒TE=-GMm2r∝mr

⇒TE1TE2=m1 m2·r2r1=43×43=169

Topic: Physics, Gravitation, JEE Main, JEE Main 2022 (27 Jul Shift 1)
Two satellites A and B move round the earth in the same orbit. The mass of A is twice the mass of B. The quantity which is same for the two satellites will be

Options:
A Speed
B Kinetic energy
C Total energy
D Potential energy

The correct answer is:
Speed

Solution:
The formula to calculate the speed of a satellite while revolving around the Earth is given by

v=GMeR ...1

where, v is the speed, G is the Gravitational constant, Me is the mass of the Earth, R is the radius of the orbit.

As can be seen from the equation, speed is independent of mass of satellite.

Hence, the speed of the satellites will be the same.

Topic: Physics, Gravitation, JEE Main, JEE Main 2023 (12 Apr Shift 1)
Two satellites A and B of masses 200 kg and 400 kg are revolving round the earth at height of 600 km and 1600 km respectively. If TA and TB are the time periods of A and B respectively then the value of TB-TA :

[ Given : radius of earth =6400 km, mass of earth =6×1024 kg ]

Options:
A 4.24×103 s
B 3.33×102 s
C 1.33×103 s
D 4.24×102 s

The correct answer is:
1.33×103 s

Solution:
⇒T=2πr3GM

⇒TA=2π6400+600×103GM

⇒TB=2π×10983GM

⇒TB-TA=2π109GM88-77

=314×4.107

=1289.64

=1.289×103 s

Topic: Physics, Gravitation, JEE Main, JEE Main 2021 (25 Feb Shift 1)
Two satellites of earth, $S_{1}$ and $S_{2}$, are moving in the same orbit. The mass of $S_{1}$ is four times the mass of $S_{2}$. Which one of the following statements is true ?

Options:
A The time period of $S_{1}$ is four times that of $S_{2}$
B The potential energies of earth and satellite in the two cases are equal
C $S_{1}$ and $S_{2}$ are moving with the same speed
D The kinetic energies of the two satellites are equal

The correct answer is:
$S_{1}$ and $S_{2}$ are moving with the same speed

Solution:
When two satellites of earth are moving in same orbit, then time period of both are equal. From Kepler's third law
$$
T^{2} \propto r^{3}
$$
Time period is independent of mass, hence their time periods will be equal. The potential energy and kinetic energy are mass dependent, hence the PE and $\mathrm{KE}$ of satellites are not equal. But, if they are orbiting in a same orbit, then they have equal orbital speed.

Topic: Physics, Gravitation, JEE Main
Two satellites of masses m and 3 m revolve around the earth in circular orbits of radii r & 3r respectively. The ratio of orbital speeds of the satellites respectively is

Options:
A 3:1
B 3:1
C 9:1
D 1:1

The correct answer is:
3:1

Solution:
The orbital velocity is given

V=GMr

where M is the mass of earth and r is the radius of the orbit of satellite.

The velocity of the satellite with mass m is

V1=GMr

and the velocity of the satellite with mass 3m is

V2=GM3r

Hence, the ratio of the velocities is

V1V2=3rr=31

Topic: Physics, Gravitation, JEE Main, JEE Main 2023 (10 Apr Shift 1)
Two satellites revolve around a planet in coplanar circular orbits in anticlockwise direction. Their period of revolutions are 1 hour and 8 hours respectively. The radius of the orbit of nearer satellite is 2×103 km. The angular speed of the farther satellite as observed from the nearer satellite at the instant when both the satellites are closest is πxradh-1, where x is _______.

Options:

The correct answer is:
3

Solution:
T1=1 hour

⇒ω1==2π rad/hour

T2=8 hours

⇒ω2=π4 rad/hour

R1=2×103 km

As T2∝R3

⇒R2R13=T2T12⇒R2R1=182/3=4⇒R=8×103 km

V1=ω1R1=4π×103 km/h

V1=ω2R2=2π×103 km/h

Relative ω=V1-V2R2-R1=2π×1036×103=π3 rad/hour

x=3

Topic: Physics, Wave Optics, JEE Main, JEE Main 2021 (01 Sep Shift 2)