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Section:
Physics
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© examsnet.com
Question : 17 of 31
Marks:
+1
,
-0
In an aluminium(Al) bar of square cross section,a square hole is drilled and is filled with iron(Fe) as shown in the figure.The electrical resistivities of Al and Fe are
2.7
×
1
0
8
Ω
m
and
1.0
×
1
0
−
7
Ω
m
2.7\times10^{8}\,\Omega\,\mathrm{m}\ \text{and}\ 1.0\times10^{-7}\,\Omega\,\mathrm{m}
2.7
×
1
0
8
Ω
m
and
1.0
×
1
0
−
7
Ω
m
,respectively.
The electrical resistance between the two faces P and Q of the composite bar is
[JEE Adv 2015 P1]
2475
64
μ
Ω
\frac{2475}{64}\,\mu\Omega
64
2475
μ
Ω
1875
64
μ
Ω
\frac{1875}{64}\,\mu\Omega
64
1875
μ
Ω
1875
49
μ
Ω
\frac{1875}{49}\,\mu\Omega
49
1875
μ
Ω
2475
132
μ
Ω
\frac{2475}{132}\,\mu\Omega
132
2475
μ
Ω
Validate
Solution:
R
F
e
=
ρ
F
e
×
50
×
1
0
−
3
(
2
×
1
0
−
3
)
2
=
1250
μ
Ω
R_{Fe}=\frac{\rho_{Fe}\times50\times10^{-3}}{(2\times10^{-3})^{2}}=1250\,\mu\Omega
R
F
e
=
(
2
×
1
0
−
3
)
2
ρ
F
e
×
50
×
1
0
−
3
=
1250
μ
Ω
R
A
l
=
ρ
A
l
×
50
×
1
0
−
3
(
49
−
4
)
×
1
0
−
6
=
30
μ
Ω
R_{Al}=\frac{\rho_{Al}\times50\times10^{-3}}{(49-4)\times10^{-6}}=30\,\mu\Omega
R
A
l
=
(
49
−
4
)
×
1
0
−
6
ρ
A
l
×
50
×
1
0
−
3
=
30
μ
Ω
R
e
q
=
1250
×
30
1280
=
1875
64
μ
Ω
R_{eq}=\frac{1250\times30}{1280}=\frac{1875}{64}\,\mu\Omega
R
e
q
=
1280
1250
×
30
=
64
1875
μ
Ω
© examsnet.com
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