(1) aAnswer
(2) d
(3) b
Consider an infinitesimally small length element dx of the conductor at a distance x from its one end. Then, the resistance dR of the length element dx of the conductor isAnswer
$dR=\rho \frac {dx}{A}= \rho _0 (a+bx) \frac {dx}{A}$
The resistance of the whole length of the conductor is
$R=\int_{0}^{L} dR= \int_{0}^{L} \rho _0 (a+bx) \frac {dx}{A} =\frac {\rho _0 }{A} (aL + \frac {bL^2}{2})$
obtained on solving the integral
We know that I = neAvd , Drift Velocity$v_d=\frac {1}{neA} \alpha \frac {1}{A}$ ,Answer
For non-uniform cross-section drift speed will be different for different situations. Only current (rate of flow of charge) will same.
We know thatAnswer
$v_d=\frac {eE}{m} \tau $
If L is the length of the copper wire and V is the potential difference across it then
$v_d=\frac {e}{m} \frac {V}{L} \tau $
$v_d \alpha V$ i.e. it potential difference is increase drift velocity of electrons also increases
Let La and Lb be the lengths of two wires.Answer
Given that ρA=ρ i.e. ρB=2ρ also diameter of wire A is DA=D then for wire B DB=2D,
The resistance of the wire made of material A,
$R_A= \frac {\rho _A L_A}{\pi (\frac {D_A}{2})^2} = \frac {\rho _A L_A}{\pi (\frac {D}{2})^2} = \frac {4\rho L_A}{\pi D^2}$
Resistance of the wire made of material B
$R_B= \frac {\rho _B L_B}{\pi (\frac {D_B}{2})^2} = \frac {\rho _B L_B}{\pi (\frac {2D}{2})^2} = \frac {2\rho L_B}{\pi D^2}$
Since RA=RB, we have
$\frac {4\rho L_A}{\pi D^2}=\frac {2\rho L_B}{\pi D^2}$
$\frac {L_A}{L_B} =\frac {1}{2} $
Resistance at the end B and C do not be the part of network of resistance between point A and D.Answer
Therefore equivalent resistance between points A and D is
$R= 10 + \frac {1}{10+10} + \frac {1}{10+10} +10$
$R=30 \Omega $
Resistance of the segment AB and BC are in series. Therefore effective resistance isAnswer
$R_1=9+5=14\Omega$
Similarly resistance of the segment AD and DC are in series. IfR2 is the effective resistance
$R_2 =3+6 =9 \Omega$
The resistanceR1 andR2 are in parallel with $7 \Omega $ resistor. Then equivalent resistance is
$\frac {1}{R_eq}= \frac {1}{14} + \frac {1}{9} + \frac {1}{7]$
Req =3.07 $\Omega$
Above network of resistors is equivalent to the arrangement shown below in the figureAnswer
If R1 is equivalent resistance of parallel combination then
$\frac {1}{R_1 = \frac {1} {3R} + \frac {1} {3R} + \frac {1} {3R}$
R1 = R
Therefore equivalent resistance of the network between A and B is
Req=R+R=2R
Consider a shell of radius r and thickness dr as shown in figureAnswer
Resistance of the shell
$dR=\frac{dr}{\sigma4\pi r^2}$
Total resistance
$R=\int_{a}^{b}\frac{dr}{\sigma4\pi r^2}$
$=\frac{1}{4\pi\sigma}(\frac{1}{a}-\frac{1}{b})$
We know that Answer
$J=\frac{I}{A}$
And
$J=\sigma E$
So we can say that
$E=\frac{I}{\sigma A}$
Now
$V=IR$
So
$E=\frac{V}{\sigma RA}$
For distance r from center
$E=\frac{V}{\sigma R4\pi r^2}$
Substituting the value of R from above solution
$E=\frac{V}{(\frac{1}{a}-\frac{1}{b})}\frac{1}{r^2}$
Drift velocity is given byAnswer
$v_d=\frac {V}{\rho l ne}$
$v_d$ does not depend upon diameter.
Hence (c) is the correct option
In a circuit, any circuit element placed between points at the same potential can be removed, without affecting the rest of the circuit, here, by symmetry, points P, Q and R are at same potential, for any potential difference between A and BAnswer
The circuit can therefore be reduced as shown below
Now this can be reduced to
Therefore Effective Resistance
$\frac {1}{R_{eq}} = \frac {1}{4R} + \frac {1}{4R} + \frac {1}{2r}$
$\frac {1}{R_{eq}} =\frac {r + r + 2R}{4Rr}$
$R_{eq}=\frac {2Rr}{R+r}$
(a)Answer
Heat gained by water = Heat supplied by container - heat lost Answer
$m s \Delta T = 1000t - 160t $
$t= \frac {2 \times 4.2 \times 1000 \time 50}{840}$ = 8 min 20 sec
(a)Answer
(c) Answer
Here,heat supplied = the change in internal energyAnswer
$Q= \Delta U$
Now heat supplied $Q=i^2 R t$
$=1 \times 1 \times 100 \times 5 \times 60=30000$ J=30kJ.
Hence (b) is the correct option
(a), (b)Answer