The resistance of a copper wire is given R=R0(1+α(θ−20)) where R0=(4.00±0.01)Ω is the resistance of wire at θ=20 C 0. The value of resistance coefficient of copper is (4×10−3±1×10−4)/ C 0. It is connected with a battery of emf (10.0±0.1)V. Choose CORRECT option(s).

At θ=25 C 0R=R0[1+α(25−20)] =4[1+4×10−3×5] =4.08Ω ΔRR=ΔR0R0+Δ[1+α(θ−20)][1+α(θ−20)]=ΔR0R0(Δα)(θ−20)[1+α(θ−20)]=0.014.00+10−4×51.02 =0.0025+4.9×10−4 ΔRR=0.0029 ΔR=0.0118 R=(4.08±0.01)α At θ=500C R=4 At θ=500C R=4(1+4×10−3×30) R=4.48ΩΔRR=ΔRR0+(Δα)(θ−20)[1+α(θ−20)]=0.014.00+10−4×51.12=2.5×10−3+2.67×10−3=5.17×10−3ΔR=0.023R=(4.48±0.02)ΩAtθ=250C,P=1004.08=24.5WΔPP=2ΔVV+ΔRR=2×0.110+0.0029 =0.02 ΔPP=2×ΔVV+ΔRR=2×0.110+5.17×10−3=0.025

The resistance of a copper wire is given R=R0(1+α(θ−20)) where R0=(4.00±0.01)Ω is the resistance of wire at θ=20 C 0. The value of resistance coefficient of copper is (4×10−3±1×10−4)/ C 0. It is connected with a battery of emf (10.0±0.1)V. Choose CORRECT option(s).

At θ=25 C 0R=R0[1+α(25−20)] =4[1+4×10−3×5] =4.08Ω ΔRR=ΔR0R0+Δ[1+α(θ−20)][1+α(θ−20)]=ΔR0R0(Δα)(θ−20)[1+α(θ−20)]=0.014.00+10−4×51.02 =0.0025+4.9×10−4 ΔRR=0.0029 ΔR=0.0118 R=(4.08±0.01)α At θ=500C R=4 At θ=500C R=4(1+4×10−3×30) R=4.48ΩΔRR=ΔRR0+(Δα)(θ−20)[1+α(θ−20)]=0.014.00+10−4×51.12=2.5×10−3+2.67×10−3=5.17×10−3ΔR=0.023R=(4.48±0.02)ΩAtθ=250C,P=1004.08=24.5WΔPP=2ΔVV+ΔRR=2×0.110+0.0029 =0.02 ΔPP=2×ΔVV+ΔRR=2×0.110+5.17×10−3=0.025

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The resistance of a copper wire is given $R = R_{0} (1 + α (θ - 20))$ where $R_{0} = (4.00 \pm 0.01) Ω$ is the resistance of wire at $θ = 20 {}^{0}C$ . The value of resistance coefficient of copper is $(4 \times 10^{- 3} \pm 1 \times 10^{- 4}) / {}^{0}C$ . It is connected with a battery of emf $(10.0 \pm 0.1) V$ . Choose CORRECT option(s).

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The resistance $R$ at $θ = 50 {}^{0}C$ is $(4.48 \pm 0.02) Ω$

The percentage error in calculation of power loss in resistance at $θ = 50 {}^{0}C$ is 4%

The resistance $R$ at $θ = 25 {}^{0}C$ is $(4.08 \pm 0.01) Ω$

The percentage error in calculation of power loss in resistance at $θ = 25 {}^{0}C$ is 2%

answer is A, B, C.

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Detailed Solution

At $θ = 25 {}^{0}C$
$R = R_{0} [1 + α (25 - 20)] = 4 [1 + 4 \times 10^{- 3} \times 5] = 4.08 Ω \frac{ΔR}{R} = \frac{{ΔR}_{0}}{R_{0}} + \frac{Δ [1 + α (θ - 20)]}{[1 + α (θ - 20)]} = \frac{{ΔR}_{0}}{R_{0}} \frac{(Δα) (θ - 20)}{[1 + α (θ - 20)]} = \frac{0.01}{4.00} + \frac{10^{- 4} \times 5}{1.02} = 0.0025 + 4.9 \times 10^{- 4} \frac{ΔR}{R} = 0.0029 ΔR = 0.0118 R = (4.08 \pm 0.01) α At θ = 50^{0} C R = 4 At θ = 50^{0} C R = 4 (1 + 4 \times 10^{- 3} \times 30) R = 4.48 Ω$
$\begin{array}{l} \frac{ΔR}{R} = \frac{ΔR}{R_{0}} + \frac{(Δα) (θ - 20)}{[1 + α (θ - 20)]} = \frac{0.01}{4.00} + \frac{10^{- 4} \times 5}{1.12} \\ = (2.5 \times 10^{- 3}) + (2.67 \times 10^{- 3}) = 5.17 \times 10^{- 3} \\ ΔR = 0.023 \\ R = (4.48 \pm 0.02) Ω \\ Atθ = 25^{0} C, P = \frac{100}{4.08} = 24.5 W \\ \frac{ΔP}{P} = \frac{2 ΔV}{V} + \frac{ΔR}{R} \\ = 2 \times \frac{0.1}{10} + 0.0029 = 0.02 \frac{ΔP}{P} = 2 \times \frac{ΔV}{V} + \frac{ΔR}{R} = 2 \times \frac{0.1}{10} + 5.17 \times 10^{- 3} = 0.025 \end{array}$