Explanation
Consider a cube as shown in the fig-1; with uniformly distributed charge within its volume. The potential at one of its vertex P is V0.A cubical portion of half the size (half edge length) of the original cube is cut and removed as shown in the fig-Find the modulus of potential at the point P in the new structure.
A potentiometer has a wire of 100 cm length and its resistance is 10 ohms. It is connected in series with a resistance of 40 ohms and a battery of emf 2 V and negligible internal resistance. If a source of unknown emf E connected in the secondary is balanced by 40 cm length of potentiometer wire, the value of E is:
For the following circuits, the potential difference betweenX and Y in volt is (Vx−Vy)
Step1: Slope of V-I graph
Ohm's Law states that V=IR
So, V1I1 =tanθ = R1
& V2I2 =tan(90−θ) =cotθ = R2
Step2: Calculation of T2−T1
Temperature dependence of resistance is given by:
R1=R0(1+αT1)
R2=R0(1+αT2)
Subtracting above equations, we get
R2−R1=α(T2−T1)
∴T2−T1 is proportional to R2−R1
⇒ T2−T1 ∝ (cotθ−tanθ)=cosθsinθ−sinθcosθ=cos2θ−sin2θsinθcosθ
T2−T1 ∝ 2 cos2θsin2θ = 2cot2θ
Hence, T2−T1 ∝ cot2θ
Therefore, Option C is correct
Length of potentiometer wire l=1m
Emf e=E1
Emf of cell E2=1.02v/s
Null point =51cm=0.51m
(a)Potential gradient of the wire
V=1.020.51
V=2V/m
(b) Emf of cell
VPQ=E1
E1=2V
(c) When switch S is closed, null point does not shifts because current if not drawn from E2 again.
A circular coil of area 8m2 and number of turns 20 is placed in a magnetic field of 2T with its plane perpendicular to it. It is rotated with an angular velocity of 20rev/s about its natural axis. The emf induced is
Given,
Resistance become R=3Ro
R=R0(1+αΔT)
3Ro=Ro(1+αΔT)
ΔT=2α=24×10−3=500oC
Hence, when rise in temperature is 500oC
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