datavorous/entrance-exam-dataset huggingface dataset:
Three resistances $P, Q, R$ each of $2 \Omega$ and an unknown resistances $\mathrm{S}$ form the four arms of a Wheatstone bridge circuit. When a resistance of $6 \Omega$ is connected in parallel ot $\mathrm{S}$ the bridge gets balanced. What is the value of $\mathrm{S}$ ?

Options:
A $3 \Omega$
B $6 \Omega$
C $1 \Omega$
D $2 \Omega \quad ।$

The correct answer is:
$3 \Omega$

Solution:
The bridge formed with given resistances is a balanced Wheatstone's bridge
The situation can be depicted as shown in the figure.

As resistances \(S\) and \(6 \Omega\) are in paralle, their effective resistance is \(\frac{6 S}{6+S} \Omega\).
For balancing condition, \(\frac{P}{Q}=\frac{R}{\left(\frac{6 S}{6+S}\right)}\)
or \(\frac{2}{2}=\frac{2(6+\mathrm{S})}{6 \mathrm{~S}}\)
or \(3 \mathrm{~S}=6+\mathrm{S} \Rightarrow \mathrm{S}=3 \Omega\)

Topic: Physics, Current Electricity, JEE Main
Three resistors of $4 \Omega, 6 \Omega$ and $12 \Omega$ are connected in parallel and the combination is connected in series with a $1.5 \mathrm{~V}$ battery of $1 \Omega$ internal resistance. The rate of Joule heating in the $4 \Omega$ resistor is

Options:
A
$0.55 \mathrm{~W}$

B
$0.33 \mathrm{~W}$

C
$0.25 \mathrm{~W}$

D
$0.86 \mathrm{~W}$

The correct answer is:
$0.25 \mathrm{~W}$

Solution:
Resistors $4 \Omega, 6 \Omega$ and $12 \Omega$ are connected in parallel, its equivalent resistance $(R)$ is given by
$$
\frac{1}{R}=\frac{1}{4}+\frac{1}{6}+\frac{1}{12} \Rightarrow R=\frac{12}{6}=2 \Omega
$$
Again $R$ is connected to $1.5 \mathrm{~V}$ battery whose internal resistance $r=1 \Omega$.
Equivalent resistance now,
$$
R^{\prime}=2 \Omega+1 \Omega=3 \Omega
$$
Current, $I_{\text {total }}=\frac{V}{R^{\prime}}=\frac{1.5}{3}=\frac{1}{2} \mathrm{~A}$
$$
\begin{aligned}
& I_{\text {total }}=\frac{1}{2}=3 x+2 x+x=6 x \\
& \Rightarrow x=\frac{1}{12}
\end{aligned}
$$
$\therefore$ Current through $4 \Omega$ resistor $=3 x$
$$
=3 \times \frac{1}{12}=\frac{1}{4} \mathrm{~A}
$$
Therefore, rate of Joule heating in the $4 \Omega$ resistor
$$
=I^2 R=\left(\frac{1}{4}\right)^2 \times 4=\frac{1}{4}=0.25 \mathrm{~W}
$$

Topic: Physics, Current Electricity, JEE Main, JEE Main 2012 (12 May Online)
Three resistors of equal resistance are arranged in four different combinations, as shown below. Arrange the combinations in increasing order of power dissipation if it is known that each combination carries the same current.

Options:
A ​III<II<IV<I
B II<III<IV<I
C I<IV<III<II
D ​I<III<II<IV

The correct answer is:
​III<II<IV<I

Solution:
P=I2R
Current is same, so P â R
In the first case, it is 3r, in second case it is 23r, in the third case it is r 3 and in the fourth case the net resistance is 3 r 2 .
RIII < RII < RIV < RI

â´ PIII<PII<PIV<PI

Topic: Physics, Current Electricity, JEE Main
Three rings, each having equal radius R, are placed mutually perpendicular to each other and each having its centre at the origin of co-ordinate system. If current I is flowing through each ring, then the magnitude of the magnetic field at the common centre is

Options:
A 3μ0I2R
B zero
C 2-1μ0I2R
D 3-2μ0I2R

The correct answer is:
3μ0I2R

Solution:
B â = B x i ^ + B y j ^ + B z k ^ â B = B x 2 + B y 2 + B z 2
Since Bx = By = Bz so B = 3 B 0
B=3Î¼0I2R

Topic: Physics, Magnetic Effects of Current, JEE Main
Three rods AB, BC and BD of same length lâ¡ and cross section A are arranged as shown. The end D is immersed in ice whose mass is 440 g and is at 0 Â°C. The end C is maintained at 100 Â°C. Heat is supplied at a constant rate of 200 cal s-1. Thermal conductivities of AB , BC and BD are K, 2K and K/2 , respectively. Time after which whole ice will melt is nÃ102 s. What is the value of n? (K=100âcal m-1s-1Â°C-1, A=10âcm2,âl=1âm)

Options:

The correct answer is:
8

Solution:
Let θ be the temperature of B, then

2 KA θ - 1 0 0 l⁡ + K 2 A θ - 0 l⁡ = 2 0 0

Substituting the values we obtain,

2θ-200+θ2=2000

θ=880 °C

Rate of heat flow from B to D

KA2(θ−0)l=0.12(880)=44 cal s-1
t= 440×80 44 =800 s

⇒n=8

Topic: Physics, Thermal Properties of Matter, JEE Main
Three rods each of length \(l\) and cross sectional area \(A\) joined in series between two heat reservoirs as shown in the figure. Their conductivities are \(2 K, K\) and \(\frac{K}{2}\), respectively. Assuming that the conductors are insulated from surroundings, the temperatures \(T_1\) and \(T_2\) of the junctions in steady state condition are respectively.

Options:
A \(\frac{600}{7}{ }^{\circ} \mathrm{C}, \frac{400}{7}{ }^{\circ} \mathrm{C}\)
B \(\frac{600}{7}{ }^{\circ} \mathrm{C}, \frac{700}{4}{ }^{\circ} \mathrm{C}\)
C \(\frac{500}{6}{ }^{\circ} \mathrm{C}, \frac{600}{5}{ }^{\circ} \mathrm{C}\)
D \(\frac{600}{4}{ }^{\circ} \mathrm{C}, \frac{400}{7}{ }^{\circ} \mathrm{C}\)

The correct answer is:
\(\frac{600}{7}{ }^{\circ} \mathrm{C}, \frac{400}{7}{ }^{\circ} \mathrm{C}\)

Solution:
Key idea In a series combination of heat conductors, the rate of heat flow remains constant.
According to the question, the figure given below shows the conduction of heat through series combinations of rods.
where, \(T_1, T_2=\) temperature of junctions,
\(I=\text { heat current }=\frac{d Q}{d t}\)
As, \(\frac{d Q_1}{d t}=\frac{d Q_2}{d t}=\frac{d Q_3}{d t}=\frac{T_i-T_f}{k_{\text {eq }}}\)
So, \(\frac{k_1 A_1\left(T_i-T_1\right)}{l_1}=\frac{k_2 A_2\left(T_1-T_2\right)}{l_2}=\frac{k_3 A_3\left(T_2-T_f\right)}{l_3}\)
\(\begin{array}{rlrl}
& \ddots & A_1 & =A_2=A_3 \\
& \text {and } & l_1 & =l_2=l_3 \\
\Rightarrow & 2 k\left(100-T_1\right) & =k\left(T_1-T_2\right)=0.5\left(T_2-0\right) \ldots (i)
\end{array}\)
Equivalent coefficient of thermal conductivity,
\(\begin{aligned}
& \frac{1}{k_{\mathrm{eq}}}=\frac{1}{2 k}+\frac{1}{k}+\frac{2}{k} \\
& \Rightarrow k_{\text {eq }}=\frac{2 k}{7} \\
& \because \text { Heat current, } \frac{d Q}{d t}=\frac{d Q_1}{d t} \\
& \Rightarrow \frac{k_{e q} A\left(T_i-T_f\right)}{l}=\frac{k_1 A_1\left(T_i-T_1\right)}{l_1}
\end{aligned}\)
here, \(l=l_1+l_2+l_3=l_1+l_1+l_1=3 l_1\) and
\(A=A_1+A_2+A_3=A_1+A_1+A_1=3 A_1\)
So, \(\frac{2 k}{7} \times \frac{3 A_1}{3 l_1}(100-0)=\frac{2 k A_1\left(100-T_1\right)}{l_1}\)
or \(\frac{1}{7} \times 100=100-T_1\)
\(\begin{array}{ll}
\Rightarrow & T_1=100-\frac{100}{7} \\
\Rightarrow & T_1=\frac{600^{\circ} \mathrm{C}}{7}
\end{array}\)
Similarly,
From Eq. (i), we get
\(\begin{aligned}
2 k\left(100-T_1\right) & =0.5 k T_2 \\
\Rightarrow 200-2 \times \frac{600}{7} & =0.5 T_2 \\
\Rightarrow T_2 & =\frac{400^{\circ}}{7} \mathrm{C}
\end{aligned}\)
Hence, the correct option is (a).

Topic: Physics, Thermal Properties of Matter, AP EAMCET, AP EAMCET 2019 (23 Apr Shift 1)
Three rods made from the same material and having the same cross-sectional area, form the sides of an isosceles triangle ABC, right-angled at B. The points A and B are maintained at temperature T and 2T, respectively. In steady-state, the temperature of the point C is TC. Assuming that only heat conduction takes place along the lengths of the rods, the value of TCT is

Options:
A 122-1
B 32+1
C 1 3 2 - 1
D 1 2 + 1

The correct answer is:
32+1

Solution:
Î Q Î t BC = Î Q Î t CA
â kA 2 T - T C a = kA T C - T 2 a
Solve to get
T C T = 3 2 + 1

Topic: Physics, Thermodynamics, JEE Main
Three rods made of same material and having same cross-section have been joined as shown in the figure. Each rod is of same length. The left and right ends are kept at \( 0^{\circ} \mathrm{C} \) and \( 90^{\circ} \mathrm{C} \) respectively. The temperature of the junction of the three rods will be

Options:
A \( 45^{0} \mathrm{C} \)
B \( 60^{\circ} \mathrm{C} \)
C \( 30^{0} \mathrm{C} \)
D \( 20^{0} \mathrm{C} \)

The correct answer is:
\( 60^{\circ} \mathrm{C} \)

Solution:
Let the temperature of junction be Q. In equilibrium, rate of flow of heat through rod 1= sum of rate of flow of heat through rods 2 and 3.
dQdt1=dQdt2+dQdt3
KAθ-0l= KA90°-θl+KA90°-θl
θ =2(90°-θ)
3θ=180° , θ=180°3=60°

Topic: Physics, Thermal Properties of Matter, JEE Main
Three rods made of the same material and having the same cross-section are joined as shown in the figure. Each rod is of the same length. What is the temperature at the junction of the three rods (in °C)?

Options:

The correct answer is:
60

Solution:
Let the temperature of function be Î¸, then
H=H1+H2
â KA(Î¸-0)L=KA(90-Î¸)L+KA(90-Î¸)L
Or Î¸=90-Î¸+90-Î¸
Or Î¸=180-2Î¸
Or 3Î¸=180
Or Î¸=60 â

Topic: Physics, Thermal Properties of Matter, JEE Main
Three rods of Copper, Brass and Steel are welded together to form a Y-shaped structure. Area of cross-section of each rod is 4 cm2. End of copper rod is maintained at 100°C. Where as ends of brass and steel are kept at 0°C. Lengths of the copper, brass and steel rods are 46, 13 and 12 cms respectively. The rods are thermally insulated from surroundings except at ends. Thermal conductivities of copper, brass and steel are 0.92, 0.26 and 0.12 CGS units respectively. Rate of heat flow through copper rod is :

Options:
A 1.2 Cal/s
B 2.4 Cal/s
C 4.8 Cal/s
D 6.0 Cal/s

The correct answer is:
4.8 Cal/s

Solution:
LCu = 46 cm; LB = 13 cm & LS=12 cm.
KCu=0.92KB=0.26KS=0.12}CGSunit
We know that R=LKA

RCu=460.92×4=12.5

RB=130.26×4=12.5
RS=120.12×4=25

Now, if T is temperature of junction then
100-TRCu=T-0RB+T-0RS
100-T12.5=T12.5+T25
∴100-T12.5=3T25
200=5T
∴T=40∘
⇒QCu=100-TRCu
=100-4012.5
=6012.5
⇒QCu=4.8 Cal/s

Topic: Physics, Thermal Properties of Matter, JEE Main, JEE Main 2014 (06 Apr)
Three rods of equal lengths are joined to form an equilateral triangle $A B C . D$ is the mid-point of $A B$. The coefficient of linear expansion is $\alpha_1$ for material of rod $A B$ and $\alpha_2$ for material of rods $A C$ and $B C$. If the distance $D C$ remains constant for small changes in temperature, then

Options:
A $\alpha_1=2 \alpha_2$
B $\alpha_1=4 \alpha_2$
C $\alpha_1=8 \alpha_2$
D $\alpha_1=\alpha_2$

The correct answer is:
$\alpha_1=4 \alpha_2$

Solution:
$C D^2=(A C)^2-(A D)^2=l^2-\left(\frac{l}{2}\right)^2$
After small change in temperature
$C D^2=(A C)^2-(A D)^2$
$=\left[l\left(1+\alpha_2 t\right)\right]^2-\left[\frac{l}{2}\left(1+\alpha_1 t\right)\right]^2$
$\therefore \quad l^2-\frac{l^2}{4}=l^2\left[1+\alpha_2^2 t^2+2 \alpha_2 t\right]$ $-\frac{l^2}{4}\left[1+\alpha_1^2 t^2+2 \alpha_1 t\right]$
Neglecting $\alpha_2^2 t^2$ and $\alpha_1^2 t^2$, (because very small quantity)
$0=l^2\left(2 \alpha_2 t\right)-\frac{l^2}{4}\left(2 \alpha_1 t\right)$
or $\quad \alpha_1=4 \alpha_2$

Topic: Physics, Thermal Properties of Matter, AP EAMCET, AP EAMCET 2010
Three rods of equal lengths are joined to form an equilateral triangle $A B C . D$ is the mid-point of $A B$. The coefficient of linear expansion is $\alpha_1$ for material of rod $A B$ and $\alpha_2$ for material of rods $A C$ and $B C$. If the distance $D C$ remains constant for small changes in temperature, then

Options:
A $\alpha_1=2 \alpha_2$
B $\alpha_1=4 \alpha_2$
C $\alpha_1=8 \alpha_2$
D $\alpha_1=\alpha_2$

The correct answer is:
$\alpha_1=4 \alpha_2$

Solution:
$C D^2=(A C)^2-(A D)^2=l^2-\left(\frac{l}{2}\right)^2$
After small change in temperature
$C D^2=(A C)^2-(A D)^2$
$=\left[l\left(1+\alpha_2 t\right)\right]^2-\left[\frac{l}{2}\left(1+\alpha_1 t\right)\right]^2$
$\therefore \quad l^2-\frac{l^2}{4}=l^2\left[1+\alpha_2^2 t^2+2 \alpha_2 t\right]$ $-\frac{l^2}{4}\left[1+\alpha_1^2 t^2+2 \alpha_1 t\right]$
Neglecting $\alpha_2^2 t^2$ and $\alpha_1^2 t^2$, (because very small quantity)
$0=l^2\left(2 \alpha_2 t\right)-\frac{l^2}{4}\left(2 \alpha_1 t\right)$
or $\quad \alpha_1=4 \alpha_2$

Topic: Physics, Thermal Properties of Matter, JEE Main
Three rods of equal lengths are joined to form an equilateral triangle $A B C . D$ is the mid-point of $A B$. The coefficient of linear expansion is $\alpha_1$ for material of rod $A B$ and $\alpha_2$ for material of rods $A C$ and $B C$. If the distance $D C$ remains constant for small changes in temperature, then

Options:
A $\alpha_1=2 \alpha_2$
B $\alpha_1=4 \alpha_2$
C $\alpha_1=8 \alpha_2$
D $\alpha_1=\alpha_2$

The correct answer is:
$\alpha_1=4 \alpha_2$

Solution:
$C D^2=(A C)^2-(A D)^2=l^2-\left(\frac{l}{2}\right)^2$
After small change in temperature
$C D^2=(A C)^2-(A D)^2$
$=\left[l\left(1+\alpha_2 t\right)\right]^2-\left[\frac{l}{2}\left(1+\alpha_1 t\right)\right]^2$
$\therefore \quad l^2-\frac{l^2}{4}=l^2\left[1+\alpha_2^2 t^2+2 \alpha_2 t\right]$ $-\frac{l^2}{4}\left[1+\alpha_1^2 t^2+2 \alpha_1 t\right]$
Neglecting $\alpha_2^2 t^2$ and $\alpha_1^2 t^2$, (because very small quantity)
$0=l^2\left(2 \alpha_2 t\right)-\frac{l^2}{4}\left(2 \alpha_1 t\right)$
or $\quad \alpha_1=4 \alpha_2$

Topic: Physics, Thermal Properties of Matter, TS EAMCET, TS EAMCET 2010
Three rods of identical cross-section and length are made of three different materials of thermal conductivity K1, K2 and K3, respectively. They are joined together at their ends to make a long rod (see figure). One end of the long rod is maintained at 100°C and the other at 0°C (see figure). If the joints of the rod are at 70°C and 20°C in steady and there is no loss of energy from the surface of the rod, the correct relationship between K1, K2 and K3 is :

Options:
A K1:K3=2:3,
K2:K3=2:5
B K1<K2<K3
C K1:K2=5:2,
K1:K3=3:5
D K1>K2>K3

The correct answer is:
K1:K3=2:3,
K2:K3=2:5

Solution:
K1(100-70)=K2(70-20)=K3(20-0)

K130=K2(50)=K3(20)

⇒K1:K2:K3=10:6:15

K1:K3=2:3

Topic: Physics, Thermal Properties of Matter, JEE Main, JEE Main 2020 (06 Sep Shift 2)
Three rods of identical cross-sectional area and made from the same metal form the sides of an isosceles triangle ABC right angled at B. The points A and B are maintained at temperatures T and 2T, respectively, in the steady-state. Assuming that only heat conduction takes place, the temperature of the point C is

Options:
A 3T2+1
B T2+1
C T32-1
D T2-1

The correct answer is:
3T2+1

Solution:
As TB > TA, heat flows from B to A through both paths BA and BCA.
Rate of heat flow in BC = Rate of heat flow in CA
KA2T-Tcl=KATc-T2l
Solving this, we get Tc=3T2+1

Topic: Physics, Thermal Properties of Matter, JEE Main
Three rods of material x and three rods of material y of identical length and cross sectional area are connected as shown in the figure. If the end A is maintained at 60∘C and the junction E at 10 ∘ C , then the temperature of junction C is [Given the thermal conductivity of x is 0.92 W/m.K and that of y is 0.46 W/m.K]

Options:
A 10 ∘  C
B 20 ∘  C
C 30 ∘  C
D 40 ∘  C

The correct answer is:
20 ∘  C

Solution:
Treating the given network of rods in terms of thermal resistance Rx and Ry with
R x = L A Ã 0.92 and R y = L A Ã 0.46 as R = L AK
so that if Rx = R, Ry = 2Rx = 2R
Now as in this bridge [(P/Q) = (R/S)], so the bridge is balanced, i.e., the temperature at junctions C and D is equal and the rod CD becomes ineffective as no heat will flow through it.
Now as the thermal resistance of the bridge between junctions B and E is
1 R BE = 1 R + R + 1 2R + 2R , i.e., R BE = 4 3 R
The total resistance of bridge between A and E will be
Req = RAB + RBE = 2R + (4/3) R = (10/3) R
So the net rate of flow of heat through the bridge will be
dQ dt = Î Î¸ R eq = 60 - 10 10/3 R = 15 R
Now if TB is the temperature at B,
[ dQ dt ] AB = Î Î¸ R AB = 60 - T B 2 R
But [ dQ dt ] AB = dQ dt , i.e., 60 - T B 2 R = 15 R , i.e., T B = 30 o C
Also at B
[ dQ dt ] AB = [ dQ dt ] BC + [ dQ dt ] BD
i.e., 15 R = 30 - T C R + 30 - T D 2R
and as TC = TD = T, 30 = 3 (30 - T),
i.e., TC = TD = T = 20o C

Topic: Physics, Thermal Properties of Matter, JEE Main
Three rods of the same length and the same area of the cross-section are joined. The temperature of two ends one T1 and T2 as shown in the figure.

As we move along the rod the variation of temperature are as shown in the following
[Rods are insulated from the surrounding except at the faces]

Options:
A K13=K23=K31
B K13=K21=K33
C K13=K23=K31
D K13=K21=K33

The correct answer is:
K13=K23=K31

Solution:
K1Atan⁡30°=K2Atan⁡45°=Κ3Atan⁡60°
⇒ K1tan⁡30°=K2tan⁡45°=K3tan⁡60°
⇒ K133=K23=K33 ⇒ K13=K23=K31

Topic: Physics, Thermal Properties of Matter, JEE Main
Three rotten apples are accidently mixed with fifteen good apples. Assuming the random variable x to be the number of rotten apples in a draw of two apples, the variance of x is

Options:
A 37153
B 57153
C 47153
D 40153

The correct answer is:
40153

Solution:
Given apples are: 3 bad, 15 good

Let X be no. of bad apples.

⇒PX=0=15C218C2=15!2!×13!18!2!×16!

⇒PX=0=15×1418×17=5×73×17

⇒PX=0=3551

⇒PX=1=3C1×15C118C2=3×159×17

⇒PX=1=517

⇒PX=2=3C218C2=39×17

⇒PX=2=3153

⇒EX=0×3551+1×517+2×3153

⇒EX=13

We know that, VarX=EX2−EX2

⇒VarX=0×3551+1×517+4×151−132

⇒VarX=1951−19

⇒VarX=40153

Topic: Mathematics, Probability, JEE Main, JEE Main 2024 (31 Jan Shift 1)
Three samples of 100 g of water (samples I, II and III), initially kept at 1 atm pressure and 298 K were given the following treatments.
Sample I was heated to 320 K and cooled to 298 K
Sample II was heated to 300 K, cooled to 273 K and heated to 298 K
Sample III was heated to 373 K and cooled to 298 K
At the end of these processes, the internal energy of

Options:
A III is the highest
B II is the highest
C I and III are the same, II is lower than that of I and III
D I, II and III are the same

The correct answer is:
I, II and III are the same

Solution:
As Internal energy change is a state function so it depends only on the initial and final temperature and does not depends on the path taken.

Topic: Chemistry, Thermodynamics (C), JEE Main
Three samples of the same gas A, B and C (Î³=32) have equal volume initially. Now, the volume of each sample is doubled. For A, the process is adiabatic; for B, it is isobaric and for C, the process is isothermal. If the final pressures are equal for all the three samples, the ratio of their initial pressures is

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

The correct answer is:
2 2 : 1 : 2

Solution:
Let PA, PB, PC be the initial pressures of the three samples and P be the final pressure of each.
For A process is adiabatic, PVγ = constant
∴ PA(V)3/2 = P (2V)3/2 ⇒ PA = 23/2 P
For B, the process is isobaric ∴ PB = P
For C, the process is isothermal, PV = constant
∴ PC(V)= P(2V) ⇒ PC= 2P
Hence, P A : P B : P C = 2 3 / 2 : 1 : 2 = 2 2 : 1 : 2

Topic: Physics, Thermodynamics, JEE Main
Three semi-infinite wires, each carrying a current of 1 A are placed on each of the coordinate axes with their ends at the origin, as shown in the figure. The magnetic induction at point P -2 m,0,0 is

Options:
A μ04πj^+k^
B μ04πj^-k^
C μ08π-j^+k^
D μ08πj^+k^

The correct answer is:
μ08π-j^+k^

Solution:
There is no magnetic field at P due to wire in x-axis BâPdue to wire in y-axis=Î¼0I4Ïak^
BâPdue to wire in z-axis Î¼0I4Ïa-j^
â´BâP=Î¼0I4Ïa-j^+k^
=Î¼0I8Ï-j^+k^=Î¼08Ï-j^+k^

Topic: Physics, Magnetic Effects of Current, JEE Main
Three solid spheres each of mass m and diameter d are stuck together such that the lines connecting the centres form an equilateral triangle of side length d. Find the ratio I0/IA, where I0 is the moment of inertia of the system about an axis passing through the centroid and IA is the moment of inertia of the system about an axis passing through the centre of any sphere and perpendicular to the plane of the triangle.

Options:
A 2313
B 1315
C 1323
D 1513

The correct answer is:
1323

Solution:
The moment of inertia of the system about the point O is

I0=325Md22+Md32=1310Md2

The moment of inertia of the system about the point A is

IA=225Md22+M(d)2+25Md22=2310Md2

So, the ratio =1323

Topic: Physics, Rotational Motion, JEE Main
Three solid spheres each of mass m and diameter d are stuck together such that the lines connecting the centres form an equilateral triangle of side of length d . The ratio I0IA of moment of inertia I0 of the system about an axis passing the centroid and about center of any of the spheres IA and perpendicular to the plane of the triangle is:

Options:
A 1323
B 1513
C 2313
D 1315

The correct answer is:
1323

Solution:
M.I about O =325Md22+Md32=1310Md2
M.I about A =225Md22+Md2+25Md22=2310Md2
Now ratio =1323

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2020 (09 Jan Shift 1)
Three solids of masses $m_1, m_2$ and $m_3$ are connected with weightless string in succession and are placed on a frictionless table. If the mass $m_3$ is dragged with a force $T$, the tension in the string between $m_2$ and $m_3$ is

Options:
A $\frac{m_2}{m_1+m_2+m_3} T$
B $\frac{m_3}{m_1+m_2+m_3} T$
C $\frac{m_1+m_2}{m_1+m_2+m_3} T$
D $\frac{m_2+m_3}{m_1+m_2+m_3} T$

The correct answer is:
$\frac{m_1+m_2}{m_1+m_2+m_3} T$

Solution:
$T^{\prime}=\left(m_1+m_2\right) \times \frac{T}{m_1+m_2+m_3}$

Topic: Physics, Laws of Motion, JEE Main
Three sound waves of equal amplitudes have frequencies $(v-1), v,(v+1)$. They superpose to give beats. The number of beats produced per second will be

Options:
A
4

B
3

C
2

D
1

The correct answer is:
2

Solution:
$\quad(3)$
Maximum number of beats $=v+1-(v-1)=2$

Topic: Physics, Waves and Sound, JEE Main, JEE Main 2009
Three sparingly soluble salts that have same solubility products as given below

I. A2X II. AX III. A2X3

Their solubilities in a saturated solution will be such that

Options:
A III > II > I
B III > I > II
C II > III > I
D II > I > III

The correct answer is:
III > II > I

Solution:
Let Ksp=10-30

10-3012=10-15

10-30413=10-10413

10-3010815=10-610815

S=Ksp for AX
S=Ksp41/3 for A2X
S=Ksp1081/5 for A2X3
So SIII>SII>SI.

Topic: Chemistry, Ionic Equilibrium, JEE Main
Three specimens A, B, C of same radioactive element has activities 1 microcurie, 1 rutherford and 1 becquerel respectively. Which specimen has maximum mass?

Options:
A A
B B
C C
D all have equal masses

The correct answer is:
A

Solution:
1 becquerel $=1 \mathrm{dis} / \mathrm{s}$

1 rutherford $=10^{6} \mathrm{dis} / \mathrm{s}$

1 curie $=3.7 \times 10^{10} \mathrm{dis} / \mathrm{s}$

Topic: Physics, Nuclear Physics, JEE Main
Three straight parallel current carrying conductors are shown in the figure. The force experienced by the middle conductor of length $25 \mathrm{~cm}$ is:

Options:
A
$3 \times 10^{-4} \mathrm{~N}$ toward right

B
$6 \times 10^{-4} \mathrm{~N}$ toward right

C
$9 \times 10^{-4} \mathrm{~N}$ toward right

D
Zero

The correct answer is:
$3 \times 10^{-4} \mathrm{~N}$ toward right

Solution:
$$
\text { } I_1=30 \mathrm{~A} \quad I=10 \mathrm{~A} \quad I_2=20 \mathrm{~A}
$$
Also given; length of wire $Q$ $=25 \mathrm{~cm}=0.25 \mathrm{~m}$
Force on wire $\mathrm{Q}$ due to wire $\mathrm{R}$
$$
\begin{aligned}
&F_{Q R}=10^{-7} \times \frac{2 \times 20 \times 10}{0.05} \times 0.25 \\
&=20 \times 10^{-5} \mathrm{~N} \text { (Towards left) }
\end{aligned}
$$
Force on wire $\mathrm{Q}$ due to wire $\mathrm{P}$
$$
\begin{aligned}
&F_{Q P}=10^{-7} \times \frac{2 \times 30 \times 10}{0.03} \times 0.25 \\
&=50 \times 10^{-5} \mathrm{~N} \text { (Towards right) } \\
&\text { Hence, } F_{\text {net }}=F_{Q P}-F_{Q R} \\
&=50 \times 10^{-5} \mathrm{~N}-20 \times 10^{-5} \mathrm{~N} \\
&=3 \times 10^{-4} \mathrm{~N} \text { towards right }
\end{aligned}
$$

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2014 (11 Apr Online)
Three unequal resistors in parallel are equivalent to a resistance 1 Î©. If two of them are in the ratio 1:2 and if no resistance value is fractional, what will be the largest of the three resistance in ohm?

Options:

The correct answer is:
6

Solution:
Let the resistances be R1, R2 and R3
â´R1R2=12â¹R1=k, R2=2k
In parallel 1R=1R1+1R2+1R3
11=1k+12k+1R3
1R3=1-1k=12k
=2k-2-12k=2k-32k
R3=2k2k-3
If k = 1, then R3 is found to be negative, which is impossible.
If k = 2, then R1=2, R2=4 R3=4
R2=R3, not satisfying the condition of the question that all resistance are unequal.
If k=3, then R1=3, R2=6
R3=2 Î©
â´ Largest resistance =6 Î©

Topic: Physics, Current Electricity, JEE Main