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JEE Advanced 2019 Paper 1
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© examsnet.com
Question : 13
Total: 54
A block of weight 100N is suspended by copper and steel wires of same cross sectional area
0.5
c
m
2
and length
√
3
m
and 1m, respectively. Their other ends are fixed on a ceiling as shown in figure. The angles subtended by copper and steel wires with ceiling are 30º and 60º, respectively. If elongation in copper wire is
(
Δ
l
c
)
and elongation in steel wire is
(
Δ
l
s
)
,then the ratio
Δ
l
c
Δ
l
s
is_______
[Young’s modulus for copper and steel are
1
×
10
11
N
/
m
2
and
2
×
10
11
N
/
m
2
respectively.]
Your Answer:
Validate
Solution:
T
S
cos
60
=
T
c
e
l
l
cos
30
T
S
2
=
T
C
u
√
3
2
T
S
=
√
3
T
C
u
T
S
sin
60
+
T
C
u
sin
30
=
100
√
3
T
C
u
×
√
3
2
+
T
C
u
2
=
100
2
T
C
u
=
100
T
C
u
=
10
T
C
u
=
50
N
T
S
=
√
3
×
50
N
⇒Elongation in Cu wire
⇒
∆
l
C
u
l
C
u
=
T
C
u
A
r
y
c
c
l
∆
l
C
u
=
l
C
u
T
C
u
A
y
C
u
→ elongation in steel wire
∆
l
s
=
l
s
T
S
A
y
S
∆
l
C
∆
l
S
=
l
C
T
C
u
A
y
S
S
l
S
T
S
A
y
stell
⇒
l
C
u
T
C
u
y
C
u
l
S
T
S
Y
stell
=
√
3
×
50
×
2
×
10
1
×
10
11
×
1
×
√
3
×
50
=
2
© examsnet.com
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