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A nichrome wire X with length (l) and cross-sectional area (A) is connected to a 10 V source and another nichrome wire Y with length (2 l) and cross-sectional area (A/2), is connected to a 20 V source - Physics

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प्रश्न

A nichrome wire X with length (l) and cross-sectional area (A) is connected to a 10 V source and another nichrome wire Y with length (2l) and cross-sectional area `(A/2)`, is connected to a 20 V source.

(a) Compare the resistances of wires X and Y. [Given that the resistivity of nichrome is (ρ).]

(b) Compare the electrical power consumed by each wire.

(c) Compare the masses of these wires. (Given that the density of nichrome is d.)

(d) State true or false:

Wire X and wire Y both show the same rise in temperature in the same time.

संख्यात्मक
चूक किंवा बरोबर
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उत्तर

Given: Length of nichrome wire X (lX) = l

Cross-sectional area of nichrome wire X (AX) = A

Potential difference across nichrome wire X (VX) = 10 V

Length of nichrome wire Y (lY) = 2l

Cross-sectional area of nichrome wire Y (AY) = `A/2`

Potential difference across nichrome wire Y (VY) = 20 V

resistivity of nichrome = ρ

(a) Let ‘RX’ and ‘RY’ be the resistance of nichrome wires X and Y, respectively.

Now, RX​ = `ρ l/A`

RY​ = `ρ (2l)/(A/2)`

= `ρ (4l)/A`

= 4RX

As, `R_Y/R_X` = 4

So, RX : RY = 1 : 4

(b) PX = `((V_X)^2)/R_X`

= `10^2/R_X`

= `100/R_X`

PY = `((V_Y)^2)/R_Y`

= `20^2/(4R_X)`

= `400/(4R_X)`

= `100/R_X`

= PX

So, PX : PY = 1 : 1

(c) Given, mass density of nichrome = d

Then, mass of wire X (mX) = d × volume = d × Al

Mass of wire Y (mY) = d × volume

= `d xx 2l xx A/2`

= d × Al

= mX

So, mX : mY = 1 : 1

(d) This statement is true.

Explanation:

Let the heat gained by the wire X be ‘QX’ and the change in temperature be ‘ΔTX’. Similarly, heat gained by the wire Y is ‘QY’ and the change in temperature is ‘ΔTY’.

Again, let the specific heat capacity of nichrome be ‘c’ and at the same time be ‘t’.

Then, QX = PX × t

= mXc × ΔTX

And QY = PY × t

= mYc × ΔTY

As, PX = PY then QX = QY for same time t.

⇒ mXc × ΔTX = mYc × ΔTY

⇒ mX × ΔTX = mY × ΔTY

So, ΔTX : ΔTY = mX : mY = 1 : 1

Hence, wire X and wire Y both show the same rise in temperature in the same time.

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