PHYS ICS

TARGET : JEE (ADVANCED)

 
Physics 

INFORMA TIO
E E ST
DPP
DAILY PRACTICE PROBLEMS

Course: VIJETA (JP), VISHWAAS (JF), VISHESH (JD) & VIJAY (JR)
NO. 3
SOLUTIONS OF DPP NO. # 3
PART–I
1. Req = 3

0.5mm
2. L.C. of the screw gauge is =  0.01mm
50
0.5mm
LØwxst dk vYirekad =  0.01mm
50
Measured diameter = 4(0.5 mm) + (0.01) (8) = 2.08
ekik x;k O;kl = 4(0.5 mm) + (0.01) (8) = 2.08
Max error in D is D = least count of screw gauge = 0.01 mm
D esa vf/kdre =kqfV D = LØwxst dk vYirekad = 0.01 mm
RD2

4
   R  D   = 1% + 2 (0.01)  100%  1% = 3 %
   2 
   m ax R D  2.08

3. 
i=
R
1 1 1 1
=  
R d bax cx
1
For minimum i, will be minimum
R
1
i U;wure gksus ds fy, Hkh U;wure gksxk
R
d 1
i.e. =0
dx R
(a  b)  c
Gives, fn;k gS x =
2
(a  b  c  4d) (5d) 5
Put imin = j[kus ij = = .
(a  b  c)d dd d

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4. Consider circuit at any time t after closing the switch :
fLop cUn djus ds i'pkr~ fdlh le; t ij ifjiFk ij fopkj djrs gSA
t
V 
I1 = and I2 = V e RC
2r R
C R
A

r r

V S
t
 V
 VA – VB = RI2 – rl1 = R × V e RC – r ×
R 2r
 t 
– V 1– 2e– RC 
 VA – VB =
2  

–V V
a= , and b =
2 2
When VB – VA = 0  t0 = RC n(2), Hence greater the R , greater the t0
tc VB – VA = 0  t0 = RC n(2), vr% R , ftruk vf/kd gksxk t0 mruk gh vf/kd gksxk
5. A A d-x
0   K 0  
2 2 x
in case fLFkfr (1)
esa Ceq
d d
0 A K 0 A
.
in case fLFkfr (2) esa C eq  dx x
0 A K 0 A

dx x
Equating these two equations. bu nksuksa lehdj.kksa dks cjkcj djus ij
2d
x=
3
A
so vr%, V1  .d
2
2d
V2  A.
3
V2 4

V1 3

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6. Let R5 is not connected between A and B
ekuk R5 , A rFkk B ds e/; tqM+k gqvk ugha gSA
VR1 VR2
R1 A R2

VA
V O

VB R4

R3 B
VR4
VR3

V – VA =  VR  VA = V –  VR = V –  V 
 R1
1 1
R
 1  R 2 

= V  R 1  R 2 – R 1  = VR 2

V
 R1  R2  R1  R 2  R1 
 1
 R2 
V – VB =  VR  VB = V –  VR
3 3

= V –  V 
R3
 R 3  R 4 
 R3  R4 – R3 
= V 
 R3  R4 
= VR 4 V

R3  R 4  R 3 
  1
 R4 
Now if R5 is connected between A and B
vc ;fn R5 , A rFkk B ds e/; tqM+k gqvk gSA
Current will flow from A to B if VA > VB
/kkjk A ls B dh vksj izokfgr gksxh ;fn VA > VB
V V R R
  3  1
 R1   R3  R 4 R 2
 1  
R
 2 R
 4
Current will flow from B to A if VA < VB
/kkjk B ls A dh vksj izokfgr gksxh ;fn VA < VB
V V R R
  1  3
 R3   R1  R 2 R 4
 1  1
R
 2 R
 2
v 1
7.  100   100 5%
v 20
 1
 100   100 4%
 25
R v  9
  
R v  100
R
 9%
R

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 10 500
8. The potential gradient = × = 0.5 Vm–1
(500  500) 10
Since potential difference across
R1 = 0.5 × 2 = 1V
Since potential difference across
R2 = 0.5 × 8 = 4V
E = 1 + 4 = 5V
Also since in meter bridge the balancing length = 20 cm.
R1 20 1
= = .
R2 80 4
10 500
foHko izo.krk × = 0.5 Vm–1
(500  500) 10
pqafd R1 ds fljksa ij foHkokUrj = 0.5 × 2 = 1V
pqafd R2 ds fljksa ij foHkokUrj = 0.5 × 8 = 4V
E = 1 + 4 = 5V
rFkk pqafd ehVj lsrq esa lUrqfyr yEckbZ = 20 cm.
R1 20 1
= = .
R2 80 4

9. At t = 0 (ij)
V0

R R

V02 3
Pconsume = = P0
2R / 3 2
After a long time : yEcs le; i'pkr
V0

R R

V 02 P
Pconsume = = 0
2R 2
Since current in B1 decreases with time so its brightness decreases.
Initially brightness of B2 is less than B1 but later on B2 will be brighter.
;gk¡ B1 esa /kkjk le; ds lkFk ?kVrh gSA vr% bldh ped ?kVrh gSA
izkjEHk esa B2 dh ped B1 ls de gksxh ijUrq ckn esa ;g vf/kd gksxhA

10. Initial izkjfEHkd

Final vfUre

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11. Current in potential in wire AB = 0.05 A
foHkoekih rkj AB esa /kkjk = 0.05 A
VAX = VXY = VYB = 1
VAY = 2V
potential difference across AP will be balance against AY.
AP ds lkis{k foHkokUrj AY ds fo:) lUrqfyr gksxkA

12.

after redrawing the circuit ifjiFk dks iqu% [khapus ds i'pkr~

(a)  4 = 5A
(b) From loop (1) ywi (1) ls
– 8(3) + E1 – 4(3) = 0   E1 = 36 volt
from loop (2) ywi (2) ls
+ 4(5) + 5(2) – E2 + 8(3) = 0
E2 = 54 volt

(c) from loop (3) ywi (3) ls

– 2R – E1 + E2 = 0
E  E1 54
R= 2 = – 36 = 9 
2 2
Ans. (a) 5.00 A (b) 36.0 V, 54.0 V (c) 9.00 

14. If we connect positive terminal of a battery to A and negative to infinity then the current through branches
will be as shown. If we connect negative to B and positive to infinity current will be in opposite direction
2
by supper -position of two situation net current in AB branch will be . Potential difference between A
3
2 2R
and B will be R . Effective resistance between A and B will be .
3 3
;fn cSVjh ds /kukRed fljs dks A ls rFkk _.kkRed fljs dks vuUr ls tksM+ ns rks bl ifjiFk dh 'kk[kkvksa esa /kkjk fp=kkuqlkj
izokfgr gksxhA ;fn cSVjh ds _.kkRed fljs dks B ls rFkk /kukRed fljs dks vuUr ls tksM+ ns rks /kkjk foifjr fn'kk esa izkokfgr
2
gksxh] rFkk bu nksuksa fLFkfr;ksa esa izokfgr /kkjk ds v/;kjksi.k }kjk 'kk[kk AB esa dqy /kkjk izokfgr gksxhA A rFkk B ds e/;
3
2 2R
foHkokUrj R gksxkA blfy, A rFkk B ds e/; rqY; izfrjks/k gksxk]
3 3

/3
/3 A /3
/6 /6
/6 /6

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15. 1 = 300  ..........(i)
–2 + 1 = 100  ..........(ii)
where,  is the potential gradient
tgk¡,  foHko izo.krk gS
2 2
 = .
1 3
16. Reading ikB~;kad 1 = 6 x 0.5 + (20 x 0.01) = 3.20 mm
Reading ikB~;kad 2 = 12 x 0.5 + (46 x 0.01) = 6.46 mm
Thickness of first rod = 3.20 – e
iryh NM+ dh eksVkbZ = 3.20 – e
Thickness of both rods = 6.46 – e
nksuks NM+ksa dh eksVkbZ = 6.46 – e
6.46 – e = 2 (3.20 – e)
 2e – e = 6.40 – 6.46  e = – 0.06 mm
17. In steady state current through capacitor will be zero.
LFkk;h voLFkk esa la/kkfj=k ls xqtjus okyh /kkjk 'kwU; gksxh
So equivalent circuit is given below.
vr% rqY; ifjiFk uhps fn;k x;k gSA
10 V
2A 1A

10 V 8V 2V

0V
18.

W B = 10 × 10 = 100 J
1
ui = × 5 × 10–6 × 102 = 250 J
2
1
uf = × 6 × 10–2 × 102 = 300 J
2
H = wB – (f – i) = 50 J.
19. For maximum error in g :
g esa vf/kdre =kqfV ds fy,
AB 1 1 1 dg dA dB
g=      
A B g A B g2 A 2 B2
dg  dA dB 
 g 2  2 
g A B 
 dg  = 2  0 .122  0 .1 25  
  4%
 g  m ax  (6) (3 ) 

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 AB
20. f=
CD
 df  dA dB d(C  D) 0.12 + 0.15  0.04  0.12
 f   A  B  CD  6 3 24 24
 
 df  = 9.67%
 f 
  m ax

21. In process of charging v = v0 (1 – e–t/) In process of discharging

vkos'ku ds nkSjku v = v0 (1 – e–t/) fujkos'ku ds nkSjku
 t =  n  V0 
 V = V0e–t/
 V0  V 

t =  n  0 
V
V
 
For the considered situation
nh xbZ fLFkfr ds fy,
    
  V0   V0 
Time taken during charging vkos'ku esa fy;k x;k le; t1 =  n    n  
  V  3V0   V  V0 
  0 
4  
0
4 
V 3V0
(from 0 to ) =    n 4   n  4   
 n 3
4 4   3 
V0 3V0
( ls rd) =    n 4   n  4   
 n 3
4 4   3 
  3V0 
  4 
Time taken during discharging fujkos'ku esa fy;k x;k le; t2 =  n    n 3
  V0  
  4  
3V0 V
(from to 0 )
4 4
3V0 V0
( ls rd)
4 4
Total time period dqy vkorZdky = 2 n 3 = 2RC n 3

22. While charging vkos'ku ds nkSjku

+ –

+q –q R

V0
q  q V0
V0  – iR  0 i
 
C RC R
CV0 3V0

When (tc) q  i
4 4R
3CV0 V0
 q i
4 4R
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 While discharging fujkos'ku ds nkSjku
+ –

+q –q R

q q
– – iR 0  
i –
C RC
3CV0 3V
 q  i – 0
4 4R
When (tc)
CV0 V
 q  i – 0
4 4R

1
23. The current through the galvanometer is ~ of total current, the S << G.
1000
1
xsYosuksehVj ls xqtjus okyh dqy /kkjk ~ gS  S << G.
1000

24. Potential difference across galvanometer = Potential difference across S.

xsYosuksehVj ij foHkokUrj = S ij foHkokUrj
 ig . G = ( – ig) . S
101 10
 10 × 10–3  10 = (1 – 10 × 10–3)  S  RS = 2
= 
1  10 99
25–27.

A A B
C D R B
A
R 2R/3

R R R
R

2R/3 2R/3 2R/3

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 V
V O

x V–x
V O
R 2R/3 R
R R
R
R

y
2R V–y
x  (v  x)
x–V + x–y + =0  10x – 2y = 5V
2/ 3
2y  v
y – v + y– x + = 0 6y – 2x = 3V
2
5V 10V 9V
 y=  , x=
7 14 14
Vx Vy
= 
R R
5V 4V 9V V 14R
= +    R AB  14
14R 14R 14R  9
C D
R


2R/3
C 2R/3 D R
R R
R
R

2R 2R/3 2R 2R/3

2R/3

10R/3
1 3 3 15 3 18
    
R eq 2R 10R 10R 10R 10R
10R 5R
R eq    R CD  5
18 9

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 A
R
A
(B) C D
C D R
2R/3 2R/3

2R/3 2R/3
B
B
2R

R
C D

8R/3
1 1 1 3 4  8  3 15
    
R CD 2R R 8R 8R 8R
8R
R CD  16
15
RAB = 1  4R  R   7R  35 
2 3  6
(C) Let RAB = xR, RCD = 2 xR
A R

R 2xR

B
R
(2xR)R
RAB = 2R +  xR
2xR  R
2x
2+ x  4x + 2 + 2x = 2x2 + x  2x2 – 5x – 2 = 0
2x  1
5  25  16 5  41
x= 
4 4
 5  41 
x=  
 4
 
 41  5   41  25 
RAB =    R   4
 4  4 
 
 41  5 
RCD = 2 
 R  8
 4 

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 PART–II
1.
0 1mm 1 2 3 4 5 6
MSD
0.5mm 0.7mm 0.9mm 1.1mm
VSD
0 1 2 3 4 5 6 7 8
1mm
0.4mm 0.6mm 0.8mm
1 VSD  0.6  0.4
2 VSD  0.1  0.9
3 VSD  0.5  0.5
4 VSD  0.8  0.2
5 VSD  0  0
6 VSD  0.1  0.9
7 VSD  0.1  0.9
8 VSD  0  0
Zero error 'kwU; =kqfV = 0 – 0.2 mm = – 0.2 mm
Reading ikB~;kad = 4 + 0.1 mm = + 4.1 mm
Diameter O;kl = (4.1 + 0.2) mm = 4.3 mm
2.
C 2q/3
+ –q C
+ –

R


+ –

C/2 q/3
q2
Let initial charge on C is q U1 = 
2C
The final charges will be in the ratio of capacitance.
C
x C  2x x + 2x = q
2
q
x=
3
2q
final charge on C = 2x =
3
C q
final charge on =x=
2 3
1 (2q / 3 ) 2 2 q2
total final energy U = + (q / 3) =
2 C 2(C / 2) 3C
loss in stored potential energy = heat generated in resistance R.
q2 q2 q2
 H = U1 – U = – =
2C 3C 6C

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final energy on the second  C  capacitor
2
1 q / 22 q2
U2 = =
2 C/ 2 9C
q2 q2 q2
 U1 : H : U2 = : :
2C 6C 9C
9:3:2

C 2q/3
+ –q C
+ –

R


+ –

C/2 q/3
q2
ekuk C ij izkjfEHkd vkos'k q U1 = 
2C
vfUre vkos'k /kkfjrk ds vuqikr esa gksxsaA
C q
x C  2x x + 2x = q ; x=
2 3
2q
C ij vfUre vkos'k = 2x =
3
C q
ij vfUre vkos'k = x =
2 3
1 2 2 q2
vfUre dqy ÅtkZ U = ( 2 q / 3 ) + (q / 3) =
2 C 2(C / 2) 3C
lafpr fLFkfrt ÅtkZ esa gkfu = izfrjks/k R esa mRiUu Å"ek
q2 q2 q2
 H = U1 – U = – =
2C 3C 6C
f}rh; la/kkfj=k  C  esa vfUre ÅtkZ
 
2
1 q / 22 q2
U2 = =
2 C/ 2 9C
q2 q2 q2
 U1 : H : U2 = : :
2C 6C 9C
9:3:2

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 40 abK
3. C=
ba
b  (b  a)
R =  dr 2 =
a 4 r 4  ab
 = RC = K0
q = q0e–t/
t

q = q 0 e  K 0
t = , q = 0
Amount of heat generated mRiUu Å"ek
2
 1 1  q0

H   
 a b  8  0 K

R R
4. <r<
6 4
5. Lets find the equivalent resistance between A and B first.
ekuk igys A rFkk B ds e/; rqY; izfrjks/k Kkr djrs gSA
Given circuit is like
fn;k x;k ifjiFk fuEu izdkj gksxkA

D R R
'' R
1
  B C D
B
' ' C R R
'
A A A B
' R

if RAB = ×R  RCD = ×R

×R

R R

1 1 1
 =
R R(x  2) x
x3 1
 =
x2 x
x2 + 3x = x + 2  x2 + 2x – 2 = 0
2  48 2  2 3
x=
2
=
2
=  3 1
RAB =  3 1 R 
Now lets work out for equivalent resistance between P and Q.
blh izdkj P rFkk Q ds e/; rqY; izfrjks/k Kkr djrs gSA
Given circuit is like
fn;k x;k ifjiFk fuEu izdkj gksxkA

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 RAB =  
3 1 R

B R R
P R
Q Q
Q A
P B R
P

A RAB =  
3 1 R

1 1 2 1  3 3 3
=  =   =
R PQ R  3 1 R R  3 1  R

R
RPQ =
3
0 5 10 15 20 25 Main scale

6.
Vernier scale

0 4 8 12 16 20
d d
2d 2d
3d 3d

d = 0.25 mm
LC = 0.25 mm
zero error 'kwU; =kqfV = – 0.75 mm
while taking the reading zero of vernier will be between 9th and 10th of main scale division.
ikB~;kad ysrs le; ofuZ;j iSekus dk 'kwU;] eq[; iSekus ds 9th rFkk 10th Hkkx ds e/; gSA
Reading = 9mm + 0.50 mm + 0.75 m = 10.25 mm
ikB~;kad = 9mm + 0.50 mm + 0.75 m = 10.25 mm

7. Equivalent Circuit
rqY; ifjiFk

1 2 2 3 3 4 4 5 5 6
P R P S
Q
Q
C1 C2 C3 C4 C5

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 P

C3 C4

R S
C1

C2 C5

0 A
C1 = C2 = C3 = C4 = C5 = C=
d
Balanced wheat stone bridge
larqfyr OghVLVksu lsrw
C C

C C

Q= CV

V
 A
W B = CV2 = 0 V 2
d
0 A V
Q3 d 2 V
Electric field between 3 & 4 (3 & 4 ds e/; fo|qr {ks=k) =
E3 
A0 A0 2d
V
Potential difference between 2 and 5 (2 & 5 ds e/; foHkokUrj) = VQ – VS =
2
V
Potential difference between 1 and 4(1 & 4 ds e/; foHkokUrj) = VP – VS =
2
  5 l 
8. (A) VAB   
 6 l    a 2 
 2 
a
5
VAB 
6
 l 2l
  3l 

(C)  ' (i)  
 2  (i)  2 
 a 2
a   6l  a 2
 2 
a

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9. Q0  q q R
– iR –  0 dq
c c i
Q0  2q Rdq +( Q0 – q) dt
 +q
c dt C1 = C C2 = C
q t –( Q0 – q) –q
dq 1
Q
0 0  2q

R c  dt
0

Q0
q
2

1 e2t /Rc 
dq Q0 2t /Rc
i  e
dt Rc
Q0
Q0 q
2
1 e2t /Rc  
Q 20
H
4c

10. (i) At t > 0 ij

i = current through dielectric ijkoS|qr ls /kkjk

q
= ...(i)
C.R.
q
By K.V.L. }kjk  – iR – =0 ...(2)
C
dq q dq
i = i + = + ...(3)
dt RC dt
q
By (2) and rFkk (3) ls  –  q  dq  R – =0
 RC dt  C
dq
 C – 2q – RC =0
dt
q t
dq dq dt
 C – 2q = RC
dt


0
C  2q
=
 RC
0

1
 – 1  n  C   2q   t  q= C(1 e2t / RC )
2  C  RC 2
C
(ii) qmax = as t  
2

and rFkk by (2) }kjk  – iR – = 0
2

  i= at that time. ml le;
2R

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 6–V
11. I = av1/2 = ....... (1)
R
6 – V 
2
Pcon = av 3/2 and rFkk PR =
R
A/c question, Pcon = 2 PR
iz'u ds vuqlkj, Pcon = 2 PR
2 6 – V 
2
3/2
= av …... (2)
R
Using (1) and (2)
(1) rFkk (2) ds mi;ksx ls
V = 2 (6 – v)
3v = 12  V = 4 Volt
2 1
2a = R= 
R a
12. The distribution of charge is shown in figure
vkos'k forj.k fp=kkuqlkj gS
q2 q q
 3  1  0
5 0.75 15
 q1 – 3q2 + 20q3 = 0 ..........(i)

 3q1 – q2 – 44q3 = 0 ...........(ii)

 q  q3  q3 q1  q 3 q
 2   0.75   3  0
 15  5 0.75
q 2  q2  q3  q2
23    5 
0
5  15 
345 = 7q2 + q3 ...........(iii)
From eq.(i), (ii), (iii) lehdj.k (i), (ii), (iii) ls
19  345 13  345 345
q1 = , q2 = , q3 =
92 92 92
q3
Potential difference between A and B = = 5V ...Ans.
0.75
q
A rFkk B ds e/; foHkokUrj = 3 = 5V ...Ans.
0.75
13. A
C1 = C5 =
d
0 A
C2 = C4 =
2d
0 A
C3 =
3d
Capacitance /kkfjrk
C1C 2 C4C5
=  C3 
C1  C 2 C4  C5
 1 1
1 1 
=
0 A  2 1
  2 =  0 A  1  1  1  = 0 A
 
d  1 3 1 d  3 3 3  d
1 1 
 2 2

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14. Given circuit can be simplified as dotted part can be replaced as

iznÙk ifjiFk ljyhd`r djus ij js[kkfdar Hkkx ds :i esa

izfrLFkkfir fd;k tk ldrk gSA
6 0

eq = 3 6 = 4V
1 1

3 6
1 1 1
=   req = 2
req 3 6
then current (rc /kkjk)
10  4 6
= =
2 R 2R
Power in R (R esa 'kfDr)
2
36R
P =  6 
 R = ,
2R  (2  R ) 2
dP
for P to be maximum =0
dR
dP
(P ds vf/kdre gksus ds fy, = 0)
dR
on solving (gy djus ij) R = 2

2   20 3
15. RAC = RCB = × = 40 
3 
2   20 6
& rFkk RAD = RBD = × = 80 
3 
120
 Balanced W.S.B lUrqfyr OghV LVksu lsrq  RCD = = 60 
2
  ×  60 
 = 48 x &  =x
 60  5 
60
  × = 48
65
   = 52 cm Ans.

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