JEE Main 2020 Physics January 9 Shift 1 Paper & Solutions | Mock Tests for JEE Main and Advanced 2025 PDF Download

 Page 1


     
 
    
    
    
     
 
                       
 
  
 
 
 
PART : PHYSICS 
 
Single Choice Type   (,dy fodYih; izdkj) 
This section contains 20 Single choice questions. Each question has 4 choices (1), (2), (3) and (4) for its 
answer, out of which Only One is correct. 
bl [k.M esa 20 ,dy fodYih iz'u gSaA izR;sd iz'u ds 4 fodYi (1), (2), (3) rFkk (4) gSa] ftuesa ls flQZ ,d lgh gSA 
 
1. Kinetic energy of the particle is E and it's De –Broglie wavelength is ?. On increasing it's KE by ?E, it's 
new De –Broglie wavelength becomes 
2
?
. Then ?E is  
 ,d d.k ftldh xfrt ÅtkZ E rFkk Mh&czkXyh rjaxnS/;Z ? gSA bldh xftt ÅtkZ esa ?E dh o`f) djus ij mldh 
u;h Mh&czksXyh rjaxnS/;Z 
2
?
 gks tkrh  gSA rc ?E gksxk&  
 (1) 3E      (2) E      (3) 2E     (4) 4E  
Ans. (1)  
Sol. ? = 
m ) KE ( 2
h
 ? ? ?
KE
1
  
 
2 / ?
?
 = 
i
f
KE
KE
 
 4KEi = KEf 
? ?E = 4KEi – KEi = 3KE = 3E    
 
2. The dimensional formula of 
G
hc
5
 is     
 
G
hc
5
 dk foeh; lw=k gksxk& 
 (1) Momentum  (2) Energy   (3) Force    (4) Pressure 
 (1) laosx   (2) ÅtkZ    (3) cy    (4) nkc 
Ans. (2) 
Sol. [ML
2
T
–2
] 
 [hc] = [ML
3
T
–2
] 
 [c] = [LT
–1
] 
 [G] = [M
–1
L
3
T
–2
]  
 
 
Page 2


     
 
    
    
    
     
 
                       
 
  
 
 
 
PART : PHYSICS 
 
Single Choice Type   (,dy fodYih; izdkj) 
This section contains 20 Single choice questions. Each question has 4 choices (1), (2), (3) and (4) for its 
answer, out of which Only One is correct. 
bl [k.M esa 20 ,dy fodYih iz'u gSaA izR;sd iz'u ds 4 fodYi (1), (2), (3) rFkk (4) gSa] ftuesa ls flQZ ,d lgh gSA 
 
1. Kinetic energy of the particle is E and it's De –Broglie wavelength is ?. On increasing it's KE by ?E, it's 
new De –Broglie wavelength becomes 
2
?
. Then ?E is  
 ,d d.k ftldh xfrt ÅtkZ E rFkk Mh&czkXyh rjaxnS/;Z ? gSA bldh xftt ÅtkZ esa ?E dh o`f) djus ij mldh 
u;h Mh&czksXyh rjaxnS/;Z 
2
?
 gks tkrh  gSA rc ?E gksxk&  
 (1) 3E      (2) E      (3) 2E     (4) 4E  
Ans. (1)  
Sol. ? = 
m ) KE ( 2
h
 ? ? ?
KE
1
  
 
2 / ?
?
 = 
i
f
KE
KE
 
 4KEi = KEf 
? ?E = 4KEi – KEi = 3KE = 3E    
 
2. The dimensional formula of 
G
hc
5
 is     
 
G
hc
5
 dk foeh; lw=k gksxk& 
 (1) Momentum  (2) Energy   (3) Force    (4) Pressure 
 (1) laosx   (2) ÅtkZ    (3) cy    (4) nkc 
Ans. (2) 
Sol. [ML
2
T
–2
] 
 [hc] = [ML
3
T
–2
] 
 [c] = [LT
–1
] 
 [G] = [M
–1
L
3
T
–2
]  
 
 
     
 
    
    
    
     
 
                       
 
  
 
 
3. Two immiscible liquids of refractive index 2 and 2 2 are filled with equal height h in a vessel. Then 
apparent depth of bottom surface of the container given that outside medium is air: 
nks v?kqyu'khy nzo ftudk viorZukad 2 rFkk 2 2 gS] dks ,d ik=k esa leku h Å¡pkbZ rd Hkjk tkrk gSA rc gok 
ls crZu ds isans dks ns[kus ij mldh vkHkklh Å¡pkbZ gksxh& 
 (1) 
4
h 2 3
     (2) 
4
h 3
     (3) 
2
h 3
    (4) 
2 4
h 3
  
Ans. (1) 
Sol.  
 
?2 = 2 2 
h 
h 
?1 = 2 
  
d = 
2 2
h
2
h
?  ? d = 
2
3
2
h
? = 
4
h 2 3
 
 
4. Three identical solid spheres each having mass 'm' and diameter 'd' are touching each other as shown 
in figure. Calculate ratio of moment of inertia about an axis (perpendicular to plane of paper) passing 
through point P and B as shown in figure. Given P is centroid of triangle ABC. 
 fn;s x;s fp=k vuqlkj rhu ,d leku Bksl xksys ftudk nzO;eku 'm' rFkk O;kl 'd' gS] dks ,d nwljs ds lEidZ esa j[kk 
tkrk gSA rks iznf'kZr fcUnq P rFkk B ls xqtjus okyh v{k ¼ry ds yEcor~½ ds lkis{k tM+Ro vk?kwZ.k dk vuqikr D;k 
gksxk ¼fn;k gS % fcUnq P, ??ABC dk dsUnzd gSA½ 
 
P 
 
A 
 
B 
 
C 
 
d 
 
 
 (1) 13 / 23    (2) 11/19    (3) 7/9     (4) 13/11  
Ans. (1) 
Page 3


     
 
    
    
    
     
 
                       
 
  
 
 
 
PART : PHYSICS 
 
Single Choice Type   (,dy fodYih; izdkj) 
This section contains 20 Single choice questions. Each question has 4 choices (1), (2), (3) and (4) for its 
answer, out of which Only One is correct. 
bl [k.M esa 20 ,dy fodYih iz'u gSaA izR;sd iz'u ds 4 fodYi (1), (2), (3) rFkk (4) gSa] ftuesa ls flQZ ,d lgh gSA 
 
1. Kinetic energy of the particle is E and it's De –Broglie wavelength is ?. On increasing it's KE by ?E, it's 
new De –Broglie wavelength becomes 
2
?
. Then ?E is  
 ,d d.k ftldh xfrt ÅtkZ E rFkk Mh&czkXyh rjaxnS/;Z ? gSA bldh xftt ÅtkZ esa ?E dh o`f) djus ij mldh 
u;h Mh&czksXyh rjaxnS/;Z 
2
?
 gks tkrh  gSA rc ?E gksxk&  
 (1) 3E      (2) E      (3) 2E     (4) 4E  
Ans. (1)  
Sol. ? = 
m ) KE ( 2
h
 ? ? ?
KE
1
  
 
2 / ?
?
 = 
i
f
KE
KE
 
 4KEi = KEf 
? ?E = 4KEi – KEi = 3KE = 3E    
 
2. The dimensional formula of 
G
hc
5
 is     
 
G
hc
5
 dk foeh; lw=k gksxk& 
 (1) Momentum  (2) Energy   (3) Force    (4) Pressure 
 (1) laosx   (2) ÅtkZ    (3) cy    (4) nkc 
Ans. (2) 
Sol. [ML
2
T
–2
] 
 [hc] = [ML
3
T
–2
] 
 [c] = [LT
–1
] 
 [G] = [M
–1
L
3
T
–2
]  
 
 
     
 
    
    
    
     
 
                       
 
  
 
 
3. Two immiscible liquids of refractive index 2 and 2 2 are filled with equal height h in a vessel. Then 
apparent depth of bottom surface of the container given that outside medium is air: 
nks v?kqyu'khy nzo ftudk viorZukad 2 rFkk 2 2 gS] dks ,d ik=k esa leku h Å¡pkbZ rd Hkjk tkrk gSA rc gok 
ls crZu ds isans dks ns[kus ij mldh vkHkklh Å¡pkbZ gksxh& 
 (1) 
4
h 2 3
     (2) 
4
h 3
     (3) 
2
h 3
    (4) 
2 4
h 3
  
Ans. (1) 
Sol.  
 
?2 = 2 2 
h 
h 
?1 = 2 
  
d = 
2 2
h
2
h
?  ? d = 
2
3
2
h
? = 
4
h 2 3
 
 
4. Three identical solid spheres each having mass 'm' and diameter 'd' are touching each other as shown 
in figure. Calculate ratio of moment of inertia about an axis (perpendicular to plane of paper) passing 
through point P and B as shown in figure. Given P is centroid of triangle ABC. 
 fn;s x;s fp=k vuqlkj rhu ,d leku Bksl xksys ftudk nzO;eku 'm' rFkk O;kl 'd' gS] dks ,d nwljs ds lEidZ esa j[kk 
tkrk gSA rks iznf'kZr fcUnq P rFkk B ls xqtjus okyh v{k ¼ry ds yEcor~½ ds lkis{k tM+Ro vk?kwZ.k dk vuqikr D;k 
gksxk ¼fn;k gS % fcUnq P, ??ABC dk dsUnzd gSA½ 
 
P 
 
A 
 
B 
 
C 
 
d 
 
 
 (1) 13 / 23    (2) 11/19    (3) 7/9     (4) 13/11  
Ans. (1) 
     
 
    
    
    
     
 
                       
 
  
 
 
Sol. M.I about P ds ifjr% tM+Ro vk?kq.kZ = 
2 2
2 d d
3 M M
52
3
??
?? ??
?? ?
?? ??
?? ?? ??
??
= 
2
Md
10
13
 
 M.I about B ds ifjr% tM+Ro vk?kq.kZ = ? ?
22
2 2 d 2 d
2 M M d M
5 2 5 2
??
? ? ? ?
?? ??
? ? ? ?
? ? ? ? ??
??
= 
2
Md
10
23
 
 Now ratio ¼vc vuqikr½ = 
13
23
 
 
5. A solid sphere having radius R and Uniform charge density ??has a cavity of radius R/2 as shown in 
figure. Find the ratio of magnitude of electric field at point A and B i.e. 
B
A
E
E
.  
 fp=kkuqlkj ,d Bksl xksyk ftldh f=kT;k R rFkk ml ij ,d leku vkos'k ?kuRo ??gSA blds vUnj R/2 f=kT;k dh 
xqgk gS] rc fcUnq A rFkk B ij fo|qr {ks=kkssa ds ifjek.k esa vuqikr  
B
A
E
E
 dk eku gksxk& 
     
 
R/2 
R 
R/2 
A 
B 
 
 (1)  
19
18
   (2) 
17
11
     (3) 
17
9
    (4) 
19
9
 
Ans. (3) 
Sol. For a solid sphere Bksl xksys ds fy, 
 E = 
0
3
r
?
?
 
 EA = 
) 3 ( 2
R
0
?
? ?
  
 
0
A
6
R
E
?
?
? 
 Electric field at point ¼fcUnq B ij fo|qr {ks=k dh rhozrk½ B = EB = E1A + E2A  
 E1A = Electric Field Due to solid sphere of radius R at point B = 
0
3
R
?
?
 
 E1A = R f=kT;k ds Bksl xksys ds dkj.k fcUnq B ij fo|qr {ks=k dh rhozrk = 
0
3
R
?
?
 
 E2A = Electric Field Due to solid sphere of radius R/2 (which having charge density – ?) 
Page 4


     
 
    
    
    
     
 
                       
 
  
 
 
 
PART : PHYSICS 
 
Single Choice Type   (,dy fodYih; izdkj) 
This section contains 20 Single choice questions. Each question has 4 choices (1), (2), (3) and (4) for its 
answer, out of which Only One is correct. 
bl [k.M esa 20 ,dy fodYih iz'u gSaA izR;sd iz'u ds 4 fodYi (1), (2), (3) rFkk (4) gSa] ftuesa ls flQZ ,d lgh gSA 
 
1. Kinetic energy of the particle is E and it's De –Broglie wavelength is ?. On increasing it's KE by ?E, it's 
new De –Broglie wavelength becomes 
2
?
. Then ?E is  
 ,d d.k ftldh xfrt ÅtkZ E rFkk Mh&czkXyh rjaxnS/;Z ? gSA bldh xftt ÅtkZ esa ?E dh o`f) djus ij mldh 
u;h Mh&czksXyh rjaxnS/;Z 
2
?
 gks tkrh  gSA rc ?E gksxk&  
 (1) 3E      (2) E      (3) 2E     (4) 4E  
Ans. (1)  
Sol. ? = 
m ) KE ( 2
h
 ? ? ?
KE
1
  
 
2 / ?
?
 = 
i
f
KE
KE
 
 4KEi = KEf 
? ?E = 4KEi – KEi = 3KE = 3E    
 
2. The dimensional formula of 
G
hc
5
 is     
 
G
hc
5
 dk foeh; lw=k gksxk& 
 (1) Momentum  (2) Energy   (3) Force    (4) Pressure 
 (1) laosx   (2) ÅtkZ    (3) cy    (4) nkc 
Ans. (2) 
Sol. [ML
2
T
–2
] 
 [hc] = [ML
3
T
–2
] 
 [c] = [LT
–1
] 
 [G] = [M
–1
L
3
T
–2
]  
 
 
     
 
    
    
    
     
 
                       
 
  
 
 
3. Two immiscible liquids of refractive index 2 and 2 2 are filled with equal height h in a vessel. Then 
apparent depth of bottom surface of the container given that outside medium is air: 
nks v?kqyu'khy nzo ftudk viorZukad 2 rFkk 2 2 gS] dks ,d ik=k esa leku h Å¡pkbZ rd Hkjk tkrk gSA rc gok 
ls crZu ds isans dks ns[kus ij mldh vkHkklh Å¡pkbZ gksxh& 
 (1) 
4
h 2 3
     (2) 
4
h 3
     (3) 
2
h 3
    (4) 
2 4
h 3
  
Ans. (1) 
Sol.  
 
?2 = 2 2 
h 
h 
?1 = 2 
  
d = 
2 2
h
2
h
?  ? d = 
2
3
2
h
? = 
4
h 2 3
 
 
4. Three identical solid spheres each having mass 'm' and diameter 'd' are touching each other as shown 
in figure. Calculate ratio of moment of inertia about an axis (perpendicular to plane of paper) passing 
through point P and B as shown in figure. Given P is centroid of triangle ABC. 
 fn;s x;s fp=k vuqlkj rhu ,d leku Bksl xksys ftudk nzO;eku 'm' rFkk O;kl 'd' gS] dks ,d nwljs ds lEidZ esa j[kk 
tkrk gSA rks iznf'kZr fcUnq P rFkk B ls xqtjus okyh v{k ¼ry ds yEcor~½ ds lkis{k tM+Ro vk?kwZ.k dk vuqikr D;k 
gksxk ¼fn;k gS % fcUnq P, ??ABC dk dsUnzd gSA½ 
 
P 
 
A 
 
B 
 
C 
 
d 
 
 
 (1) 13 / 23    (2) 11/19    (3) 7/9     (4) 13/11  
Ans. (1) 
     
 
    
    
    
     
 
                       
 
  
 
 
Sol. M.I about P ds ifjr% tM+Ro vk?kq.kZ = 
2 2
2 d d
3 M M
52
3
??
?? ??
?? ?
?? ??
?? ?? ??
??
= 
2
Md
10
13
 
 M.I about B ds ifjr% tM+Ro vk?kq.kZ = ? ?
22
2 2 d 2 d
2 M M d M
5 2 5 2
??
? ? ? ?
?? ??
? ? ? ?
? ? ? ? ??
??
= 
2
Md
10
23
 
 Now ratio ¼vc vuqikr½ = 
13
23
 
 
5. A solid sphere having radius R and Uniform charge density ??has a cavity of radius R/2 as shown in 
figure. Find the ratio of magnitude of electric field at point A and B i.e. 
B
A
E
E
.  
 fp=kkuqlkj ,d Bksl xksyk ftldh f=kT;k R rFkk ml ij ,d leku vkos'k ?kuRo ??gSA blds vUnj R/2 f=kT;k dh 
xqgk gS] rc fcUnq A rFkk B ij fo|qr {ks=kkssa ds ifjek.k esa vuqikr  
B
A
E
E
 dk eku gksxk& 
     
 
R/2 
R 
R/2 
A 
B 
 
 (1)  
19
18
   (2) 
17
11
     (3) 
17
9
    (4) 
19
9
 
Ans. (3) 
Sol. For a solid sphere Bksl xksys ds fy, 
 E = 
0
3
r
?
?
 
 EA = 
) 3 ( 2
R
0
?
? ?
  
 
0
A
6
R
E
?
?
? 
 Electric field at point ¼fcUnq B ij fo|qr {ks=k dh rhozrk½ B = EB = E1A + E2A  
 E1A = Electric Field Due to solid sphere of radius R at point B = 
0
3
R
?
?
 
 E1A = R f=kT;k ds Bksl xksys ds dkj.k fcUnq B ij fo|qr {ks=k dh rhozrk = 
0
3
R
?
?
 
 E2A = Electric Field Due to solid sphere of radius R/2 (which having charge density – ?) 
     
 
    
    
    
     
 
                       
 
  
 
 
 E2A = R/2 f=kT;k ds Bksl xksys (ftldk vkos'k ?kuRo –??gS) ds dkj.k fcUnq B ij fo|qr {ks=k dh rhozrk 
 = – 
2
R 9
4 ' KQ ?
 = –
0
54
R
?
?
 
 EB =  E1A + E2A  =  
0
3
R
?
?
 – 
0
54
R
?
?
 = 
0
54
R 17
?
?
 
 
17
9
E
E
B
A
? 
 
6. Consider an infinitely long current carrying cylindrical straight wire having radius 'a'. Then the ratio of 
magnetic field at distance 
3
a
 and 2a from axis of wire is. 
 ,d vUkUr yEcs /kkjkokgh csyukdkj lh/ks rkj fd f=kT;k 'a' gSA rc v{k ls 
3
a
 rFkk 2a nwjh ij pqEcdh; {ks=k dk 
vuqikr gksxk& 
 (1) 
5
3
     (2) 
3
2
     (3) 
2
1
     (4) 
3
4
  
Ans. (2) 
Sol.  
 
2a 
i 
A 
a/3 
B 
 
 BA = 
2
0
a 2
ir
?
?
 = 
2
0
a 2
3
a
i
?
?
 = 
2
0
a
i
?
?
6
a
 = 
a 6
i
0
?
?
  
 BB = 
) a 2 ( 2
i
0
?
?
 
 
B
A
B
B
 = 
6
4
 = 
3
2
  
 
Page 5


     
 
    
    
    
     
 
                       
 
  
 
 
 
PART : PHYSICS 
 
Single Choice Type   (,dy fodYih; izdkj) 
This section contains 20 Single choice questions. Each question has 4 choices (1), (2), (3) and (4) for its 
answer, out of which Only One is correct. 
bl [k.M esa 20 ,dy fodYih iz'u gSaA izR;sd iz'u ds 4 fodYi (1), (2), (3) rFkk (4) gSa] ftuesa ls flQZ ,d lgh gSA 
 
1. Kinetic energy of the particle is E and it's De –Broglie wavelength is ?. On increasing it's KE by ?E, it's 
new De –Broglie wavelength becomes 
2
?
. Then ?E is  
 ,d d.k ftldh xfrt ÅtkZ E rFkk Mh&czkXyh rjaxnS/;Z ? gSA bldh xftt ÅtkZ esa ?E dh o`f) djus ij mldh 
u;h Mh&czksXyh rjaxnS/;Z 
2
?
 gks tkrh  gSA rc ?E gksxk&  
 (1) 3E      (2) E      (3) 2E     (4) 4E  
Ans. (1)  
Sol. ? = 
m ) KE ( 2
h
 ? ? ?
KE
1
  
 
2 / ?
?
 = 
i
f
KE
KE
 
 4KEi = KEf 
? ?E = 4KEi – KEi = 3KE = 3E    
 
2. The dimensional formula of 
G
hc
5
 is     
 
G
hc
5
 dk foeh; lw=k gksxk& 
 (1) Momentum  (2) Energy   (3) Force    (4) Pressure 
 (1) laosx   (2) ÅtkZ    (3) cy    (4) nkc 
Ans. (2) 
Sol. [ML
2
T
–2
] 
 [hc] = [ML
3
T
–2
] 
 [c] = [LT
–1
] 
 [G] = [M
–1
L
3
T
–2
]  
 
 
     
 
    
    
    
     
 
                       
 
  
 
 
3. Two immiscible liquids of refractive index 2 and 2 2 are filled with equal height h in a vessel. Then 
apparent depth of bottom surface of the container given that outside medium is air: 
nks v?kqyu'khy nzo ftudk viorZukad 2 rFkk 2 2 gS] dks ,d ik=k esa leku h Å¡pkbZ rd Hkjk tkrk gSA rc gok 
ls crZu ds isans dks ns[kus ij mldh vkHkklh Å¡pkbZ gksxh& 
 (1) 
4
h 2 3
     (2) 
4
h 3
     (3) 
2
h 3
    (4) 
2 4
h 3
  
Ans. (1) 
Sol.  
 
?2 = 2 2 
h 
h 
?1 = 2 
  
d = 
2 2
h
2
h
?  ? d = 
2
3
2
h
? = 
4
h 2 3
 
 
4. Three identical solid spheres each having mass 'm' and diameter 'd' are touching each other as shown 
in figure. Calculate ratio of moment of inertia about an axis (perpendicular to plane of paper) passing 
through point P and B as shown in figure. Given P is centroid of triangle ABC. 
 fn;s x;s fp=k vuqlkj rhu ,d leku Bksl xksys ftudk nzO;eku 'm' rFkk O;kl 'd' gS] dks ,d nwljs ds lEidZ esa j[kk 
tkrk gSA rks iznf'kZr fcUnq P rFkk B ls xqtjus okyh v{k ¼ry ds yEcor~½ ds lkis{k tM+Ro vk?kwZ.k dk vuqikr D;k 
gksxk ¼fn;k gS % fcUnq P, ??ABC dk dsUnzd gSA½ 
 
P 
 
A 
 
B 
 
C 
 
d 
 
 
 (1) 13 / 23    (2) 11/19    (3) 7/9     (4) 13/11  
Ans. (1) 
     
 
    
    
    
     
 
                       
 
  
 
 
Sol. M.I about P ds ifjr% tM+Ro vk?kq.kZ = 
2 2
2 d d
3 M M
52
3
??
?? ??
?? ?
?? ??
?? ?? ??
??
= 
2
Md
10
13
 
 M.I about B ds ifjr% tM+Ro vk?kq.kZ = ? ?
22
2 2 d 2 d
2 M M d M
5 2 5 2
??
? ? ? ?
?? ??
? ? ? ?
? ? ? ? ??
??
= 
2
Md
10
23
 
 Now ratio ¼vc vuqikr½ = 
13
23
 
 
5. A solid sphere having radius R and Uniform charge density ??has a cavity of radius R/2 as shown in 
figure. Find the ratio of magnitude of electric field at point A and B i.e. 
B
A
E
E
.  
 fp=kkuqlkj ,d Bksl xksyk ftldh f=kT;k R rFkk ml ij ,d leku vkos'k ?kuRo ??gSA blds vUnj R/2 f=kT;k dh 
xqgk gS] rc fcUnq A rFkk B ij fo|qr {ks=kkssa ds ifjek.k esa vuqikr  
B
A
E
E
 dk eku gksxk& 
     
 
R/2 
R 
R/2 
A 
B 
 
 (1)  
19
18
   (2) 
17
11
     (3) 
17
9
    (4) 
19
9
 
Ans. (3) 
Sol. For a solid sphere Bksl xksys ds fy, 
 E = 
0
3
r
?
?
 
 EA = 
) 3 ( 2
R
0
?
? ?
  
 
0
A
6
R
E
?
?
? 
 Electric field at point ¼fcUnq B ij fo|qr {ks=k dh rhozrk½ B = EB = E1A + E2A  
 E1A = Electric Field Due to solid sphere of radius R at point B = 
0
3
R
?
?
 
 E1A = R f=kT;k ds Bksl xksys ds dkj.k fcUnq B ij fo|qr {ks=k dh rhozrk = 
0
3
R
?
?
 
 E2A = Electric Field Due to solid sphere of radius R/2 (which having charge density – ?) 
     
 
    
    
    
     
 
                       
 
  
 
 
 E2A = R/2 f=kT;k ds Bksl xksys (ftldk vkos'k ?kuRo –??gS) ds dkj.k fcUnq B ij fo|qr {ks=k dh rhozrk 
 = – 
2
R 9
4 ' KQ ?
 = –
0
54
R
?
?
 
 EB =  E1A + E2A  =  
0
3
R
?
?
 – 
0
54
R
?
?
 = 
0
54
R 17
?
?
 
 
17
9
E
E
B
A
? 
 
6. Consider an infinitely long current carrying cylindrical straight wire having radius 'a'. Then the ratio of 
magnetic field at distance 
3
a
 and 2a from axis of wire is. 
 ,d vUkUr yEcs /kkjkokgh csyukdkj lh/ks rkj fd f=kT;k 'a' gSA rc v{k ls 
3
a
 rFkk 2a nwjh ij pqEcdh; {ks=k dk 
vuqikr gksxk& 
 (1) 
5
3
     (2) 
3
2
     (3) 
2
1
     (4) 
3
4
  
Ans. (2) 
Sol.  
 
2a 
i 
A 
a/3 
B 
 
 BA = 
2
0
a 2
ir
?
?
 = 
2
0
a 2
3
a
i
?
?
 = 
2
0
a
i
?
?
6
a
 = 
a 6
i
0
?
?
  
 BB = 
) a 2 ( 2
i
0
?
?
 
 
B
A
B
B
 = 
6
4
 = 
3
2
  
 
     
 
    
    
    
     
 
                       
 
  
 
 
 
7. Find current in the wire BC. 
 rkj BC esa /kkjk Kkr djks& 
 
B 
C 
1 ? 2 ? 
4 ? 
3 ? 
20 
 
 (1) 1.6A   (2) 2A     (3) 2.4A    (4) 3A  
Ans. (2) 
Sol.  
 
B 
C 
1 ? 2 ? 
4 ? 
3 ? 
20 
4i/5 
3i/5 
2i/5 i/5 
? ?
i 
 
Reff = 
46
55
? = 2 ?  
 i = 
20
2
 
 = 10A 
 I = 
4i 3i i
–
5 5 5
?? = 2A