Physics Test - 14

The force on the positive charge moving in the magnetic field is given by Fleming's left hand rule which states that if we stretch the thumb, the centre finger and the middle finger of our left hand such that they are mutually perpendicular to each other, the centre finger gives the direction of current and middle finger points in the direction of magnetic field, then the thumb points towards the direction of the force or motion of the conductor. Direction of the moving electron is opposite to the direction of the thumb. Here, the electron is moving in the positive direction and the magnetic field is directed in the positive y-direction, so the force on the electron will be directed in negative z-direction.

Refer to Fig. Let v be the frequency of rotation.
The time taken for I full rotation is T = 1/V.
Therefore, rate of change of area is

Since the same emf is produced between the ends of each spoke, and these emfs are in parallel as is evident from Fig. The net emf between the axle and the rim of the wheel will be the same as that across each spoke. We notice that all the eight spokes are connected with one end at the rim and the other at the axle. Hence the magnitude of the net emf between the axle and the rim is independent of the number of spokes.

The velocity of transverse wave along a stretched string is given by,

$V=\sqrt{\frac{T}{P}}$

Where T is the tension and p is linear mass density since the two wires are of the same material and same radius p is constant.
So, the velocity of transverse wave depends on the tension in fact.

$V=\sqrt{T}$

Now, the ratio of transverse wave velocities is,

$\frac{V_1}{V_2}=\sqrt{\frac{T_1}{T_2}}$

$\frac{V_1}{V_2}=\sqrt{\frac{10}{20}}=\frac{1}{\sqrt{2}}$

S_n = u + a/2 (2n - 1),

S₄ = 0 + 5 (2 x 4 - 1), S₄ = 35 m

Force F on a wire of length l = iLB

Therefore, force per unit length of the wire = F/L = iB = 10^–4 x 10 = 10^–3 Nm^–1

Coefficient of performance of the refrigerator is given by

Three capacitors are connected in series, hence their equivalent capacity will be:

Cs= C / 3 = 5 / 3.

This equivalent series capacity is then connected in parallel with the fourth capacitor, hence the final equivalent capacity will be:

Cs+ C4= 5 / 3 + 5 = 20 / 3 = 6.66 6 μF.

Now, charge on C1, C2and C3is same as they are in series.

It is given by: Q = CsV = 5 / 3 x 240 = 5 x 80 x 10- 6C = 4 x 10-4C

Charge on C4= C4x V = 5 x 240 = 1200 x 10-6= 12 x 10-4C.

Given V$=\left(2\times {10}^{5}i\right)m{s}^{-1}.$The given force vector is given by

$F=q\left(V\times B\right)$

$=q\left\{2\times {10}^{5}i\times \left(i-4j-3k\right)\right\}$

$=2\times {10}^5\times q\left(-4k+3j\right)$

Therefore, the y and z components of the force are

$F_y=6\times {10}^5\times q$

and$F_z=-8\times {10}^5\times q$

$\therefore Magnitudeofforce=\sqrt{{F_y}^2+{F_z}^2}$

$=q\sqrt{{\left(6\times {10}^5\right)}^2+{\left(-8\times {10}^5\right)}^2}$

$=q\times 10\times {10}^5$

$=1.6\times {10}^{-19}\times 10\times {10}^5$

$=1.6\times {10}^{-13}N,$which is choice (3).

As Retardation is directly proportional to displacement.

$a\infty -x\Rightarrow a=-kx$
$F=-Kmx$

Using work energy theorem
Work done - change in the Kinetic energy
$dW=Fdx=-kmxdx$

Total work done
$w=|dW=|Fdx=-|kmxdx$

$W=-Km\frac{x^2}{2}$