16. Two pith balls carrying equal charges are suspended from a common point by strings of equal length. The equilibrium separation between them is r. Now the strings are rigidly clamped at half the height. The equilibrium separation between the balls now become​​
    

1. 1. 		r
      		3 √2

   
   
   

   2. 		2r
      		√3

   
   
   

   3. 		2r
      		3

   
   
   

   4. 		r			2
      		√2

   
   
   

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1. View Hint View Answer Discuss in Forum

   
   

   From figure,  tan θ =	Fe	⇒	r / 2	=	kq2
   	mg		y		r2 / mg

   
   

   [ ∵ F =	kq2	from coulomb’s law]
   	r2

   
   

   ⇒ r3 ∝ y ⇒ r'3 ∝	y	⇒	r'	=	1
   	2		r		21 / 3

   
   

   ⇒ r' =	r
   	3√2

   Correct Option: A

   
   

   From figure,  tan θ =	Fe	⇒	r / 2	=	kq2
   	mg		y		r2 / mg

   
   

   [ ∵ F =	kq2	from coulomb’s law]
   	r2

   
   

   ⇒ r3 ∝ y ⇒ r'3 ∝	y	⇒	r'	=	1
   	2		r		21 / 3

   
   

   ⇒ r' =	r
   	3√2

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17. Two identical charged spheres suspended from a common point by two massless strings of lengths l, are initially at a distance d (d << l) apart because of their mutual repulsion. The charges begin to leak from both the spheres at a constant rate. As a result, the spheres approach each other with a velocity v. Then v varies as a function of the distance x between the spheres, as :
    
    
    

1. 1. v ∝ x^{1 / 2}

   
   
   

   2. v ∝ x

   
   
   

   3. v ∝ x^{-1 / 2}

   
   
   

   4. v ∝ x^-1

   
   
   

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1. View Hint View Answer Discuss in Forum

   From figure tan θ =	Fc	; θ
   	mg

   
   

   	kq2	=	x
   	x2 mg		2ℓ

   
   
   or​ x³ ∝ q²  …(1) ​
   or​ x^{3/ 2} ∝ q …(2) ​
   Differentiating eq. (1) w.r.t. time
   

   3x2	dx	∝ 2q	dq	but	dq	is constant ​
   	dt		dt		dt

   
   so x²(v) ∝ q Replace q from eq. (2) ​
   x²(v) ∝ x^3/2 or v ∝ x^–1/2

   Correct Option: C

   From figure tan θ =	Fc	; θ
   	mg

   
   

   	kq2	=	x
   	x2 mg		2ℓ

   
   
   or​ x³ ∝ q²  …(1) ​
   or​ x^{3/ 2} ∝ q …(2) ​
   Differentiating eq. (1) w.r.t. time
   

   3x2	dx	∝ 2q	dq	but	dq	is constant ​
   	dt		dt		dt

   
   so x²(v) ∝ q Replace q from eq. (2) ​
   x²(v) ∝ x^3/2 or v ∝ x^–1/2

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18. Suppose the charge of a proton and an electron differ slightly. One of them is – e, the other is (e + ∆e). If the net of electrostatic force and gravitational force between two hydrogen atoms placed at a distance d (much greater than atomic size) apart is zero, then ∆e is of the order of [Given mass of hydrogen mh = 1.67 × 10^–27 kg]
    

1. 1. ​10^–23 C

   
   
   

   2. 10^–37 C

   
   
   

   3. 10^–47 C

   
   
   

   4. 10^–20 C

   
   
   

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1. View Hint View Answer Discuss in Forum

   According to question, the net electrostatic force (F_E) = gravitational force  (F_G)
   ​F_E = F_G
   

   or	1		∆e2	=	Gm2
   	4πε0		d2		d2

   
   

   ⇒ ∆e = m
   		G
   		K

   
   

   ∵		1	= k = 9 × 109
   		4πε0

   
   

   = 1.67 × 10–27
   		6.67 × 10–11
   		9 × 109

   
   ∆e ≈ 1.436 × 10^–37 C

   Correct Option: B

   According to question, the net electrostatic force (F_E) = gravitational force  (F_G)
   ​F_E = F_G
   

   or	1		∆e2	=	Gm2
   	4πε0		d2		d2

   
   

   ⇒ ∆e = m
   		G
   		K

   
   

   ∵		1	= k = 9 × 109
   		4πε0

   
   

   = 1.67 × 10–27
   		6.67 × 10–11
   		9 × 109

   
   ∆e ≈ 1.436 × 10^–37 C

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19. The electric field in a certain region is acting radially outward and is given by E = Aa. A charge contained in a sphere of radius 'a' centred at the origin of the field, will be given by​​
    

1. 1. A ε₀ a²

   
   
   

   2. 4 πε₀ Aa³

   
   
   

   3. ε₀ Aa³

   
   
   

   4. 4 πε₀ Aa²

   
   
   

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1. View Hint View Answer Discuss in Forum

   Net flux emmited from a spherical surface of radius a according to Gauss’s theorem
   

   φnet =	qin
   	ε0

   
   

   or,​(Aa) (4πa2) =	qin
   	ε0

   
   So, ​q_in = 4πε₀ A a³
   

   Correct Option: B

   Net flux emmited from a spherical surface of radius a according to Gauss’s theorem
   

   φnet =	qin
   	ε0

   
   

   or,​(Aa) (4πa2) =	qin
   	ε0

   
   So, ​q_in = 4πε₀ A a³
   

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20. A charge Q is placed at the corner of a cube. The electric flux through all the six faces of the cube is
    

1. 1. Q / 3ε₀

   
   
   

   2. Q / 6ε₀

   
   
   

   3. Q / 8ε₀

   
   
   

   4. Q / ε₀

   
   
   

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1. View Hint View Answer Discuss in Forum

   According to Gauss's theorem,
   

   electric flux through a closed surface =	Q	where q is charge enclosed
   	ε0

   
   by the surface.

   Correct Option: D

   According to Gauss's theorem,
   

   electric flux through a closed surface =	Q	where q is charge enclosed
   	ε0

   
   by the surface.

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