Physics: Topic-wise Test- 4 - NEET MCQ

A.Statement A is wrong.
B.m= m0​​/√(1− v2/c2) ​​
specific charge q/m=q/m0√(1​−(v2/C2))​​
so, as v increases the specific charge decreases
C. None of the fundamental particles that form the Standard Model have charge without mass. The Standard Model describes all energy and matter that makes up the universe, except gravitation. And gravitation does not have charge, it has mass.
D.Yes. Repulsion is observed only when two bodies have like charges that means the bodies must be charged. Therefore, repulsion is sure test for electrification than attraction.
 

The charge in the middle experiences force along the line. The equilibrium is stable along the line connecting charges while The equilibrium is unstable along the line perpendicular to the line of charges  
∴ Only option B is correct (given consider equilibrium only along line joining charges).
 

If field is straight then the charge moves along the straight line .if the fielss is a curve then the electrostatic force is not sufficient to change the direction along the curved field line .

There for the potential at all the points on the axis will be zero.

The acceleration due to gravity is insignificant compared with the acceleration caused by the electric field, so the gravitational field can be ignored. Hence, when a particle of mass m and charge q is thrown in a region where uniform gravitational field and electric field are present, before entering the electric field, the charge follows a straight line path (since, no net force due to any of the field). As soon as it enters the field, the charge begins to follow a parabolic path (since, constant force is always in the same direction). As soon as it leaves the field, the charge follows a straight line path  (since, no net force due to any of the field). The path of a particle may be straight line or may be parabola.

A satellite is in a state of free fall & hence weightlessness. Thus only electric force is responsible for the Tension

The dipole will have some distance along the electric field, so, option (1) is correct.

Suppose that at point B, where net electric field is zero due to charges 8q and 2q

According to conditional E_BO+ E_BA = 0

So a = L

Thus, at distance 2L from origin, net electric field will be zero       

Combination of Capacitors in Series
C(effective) = C/3 = 1

When equal charges are given to two spheres of different radii, the potential will be more or the smaller sphere as per the equation, Potential = Charge / Radius.
Since potential is inversely proportional to radius, the smaller radius will have higher potential and vice versa.

Mica capacitor. Mica capacitors have low resistive and inductive components associated with it. Hence, they have high Q factor and because of high Q factor their characteristics are mostly frequency independent, which allows this capacitor to work at high frequency.

C = Q/V
series connection splits the battery potential, hence....
V = V₁ + V₂
V₁ = Q/C₁ & V₂ = Q/C₂
Therefore, both have equal charge
 

4μf and 6μf are in series (right side of AB) so 6x(4/6)+4=24/10μf
And now 5 μf and the resulting of above two are in parallel (as on different sides of AB) so 5+(24/10) =50+(24/10)=74/10=37/5μf

For series combination,
1/C1=(1/C)+(1/C)+(1/C) [ Since all capacitors have equal capacitances]=3/C Or
C1=C/3−−−(i)
For parallel combination,
C2=C+C+C=3C−−−(ii)
Now,C1/C2=(C/3)/3C=(C/3)×(1/3C)=1/9

C1= 20×10µf

and C2= 30×10µf

in series Ceq = C1C2/(C1+C2)

Ceq = 20×10^(-6)×30×10^(-6)/20×10^(-6)+30^×10(-6)

Ceq= 12×10^(-6)f

As we know that Q = CV

Putting the values of C and V= 40V, we get

Q = (12 * 10^-6) * 40

= 480µC

Let C_p be the equivalent capacitance of parallel
C_p=C₁+C₂
16= C₁+C₂……..(1)
Let Cs be the equivalent of capacitance in series
1/C_s=(1/C₁) + (1/C₂)
Or, C_s=C₁C₂/C₁+C₂
Or,3=C₁C₂/16
Or,C₂C₁=48
C₂=48/C₁……..(ii)
16=C1+48/C1
(C₁²-16C₁+48)=0
(C₁-4)(C₁-12)=0
C₁=4 ; C₁=12
If,
C₁=4
C₂=12
If,
C₂=4
C₁=12
So, the answer is, Either 12μf or4μf

Given that some charge is given to a conductor then the whole charge is distributed over its surface only. Inside the conductor, the electric field is zero whereas potential is the same as on the surface. Hence, throughout the conductor, potential is the same i.e, the whole conductor is equipotential.

Since the electric field inside the conductor is zero and has no tangential component on its surface, therefore, no work is done in moving a test charge within the conductor or on its surface. It means the potential difference between any two points inside or on the surface is zero. Hence, electrostatic potential is constant throughout the volume of the charged conductor and has the same value on its surface as inside it.

On an irregularly shaped conductor, the surface charge density is greatest at the locations where the radius of curvature of the surface is smallest i.e, charge density is inversely proportional to radius of curvature of surface. Since the radius of curvature is least at sharp portion and maximum at flat portion, so, charge density is high at sharp portions and less at flat portions.

When there is potential difference across the conductor, the electric field is set up. Due to this charge will flow across the conductor. But when conductors have the same potential the charge will not flow from one conductor to the other.

The bridge will be balanced when the shunted resistance is of the value of 3Ω

For balanced Wheatstone's bridge
P/Q = R/S.....(i)
Power dissipation in resistance R with voltage V is V²/R.
∴ 
From equation (i)

⇒ 
Using (i), we get

∴ 

J= {J0[(x/R)-1] 0 ≤ x ≤ R/2,     {J0(x/R) R/2 ≤ x ≤ R
 We know that,
J=current/area
⇒ current=J Area
Current= J1A1 +J2A2
=J0{(x/R)−1}x(2πxdx)+J0 (x/R)(2πxdx)
=∫0R/2J0{(x/R)-1} x (2πxdx) + =∫0R/2J0 x 2π (x2/R)dx
= J0×2π∫0R/2 {(x2/R)−x}dx + (J0×2π/R) ∫RR/2 x2dx
⇒ i=J0×2π [(x3/3R)-(x2/2)] R/2 + (J0×2π/R) [x3/3]R2/R
⇒i=J0×2π [(R2/24)−(R2/8)]+ (J0×2π/R)[(R3/3)-(R3/24)]
=J0×2π(−2R2/24)(J0×2π/R)[7R3/24]
i =(5/12)πJ0R2
 

When a resistor is added parallel to other resistor in circuit it results in the reduction of overall resistance, since there are multiple pathways by which charge can flow. adding another resistor in a separate branch provides another pathway by which to direct charge through the main area of resistance within the circuit. This decreased resistance resulting from increasing the number of branches will have the effect of increasing the rate at which charge flows i.e. the current.