Question

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PLZ SOLVE Q No. 6 ​

Answer 1

ANSWER

If the answer is heat = k (I)2 * R * t -5............hre's the solution:

dimensional formula of heat(unit = joule) = [ML2T-2]

let the dependence of heat on the given 3 quantities be as follows:

[ML2T-2] = Ax * Ry * tz

where A = ampere = current, R = resistence , t = time an x, y, z their powers respectively.

equating:

[A0ML2T-2] = Ax *[ML2A-2T3]y * [T]z

= [ Ax-2y * My * L2y * T3y+z]

equating powers of same quantities on both sides:

for 'A' , .........0 = x-2y

.....................x=2y

for M,........y=1

.................x = 2

for T........3y+z = -2

................z = -2-3

................z = -5

therefore,,,

H = k (I)2 * R * t -5

Answer 2

$\sf{\underline{We\:know\:that}}$:

$\boxed{\sf{Heat\:energy = Power \times Time}}$

$\sf{\underline{That\:is}}$:

$\implies$ $\sf{I^{2} R t = \frac{V^{2}t}{R}}$

$\implies$ $\sf{H=I^{k}R^{m}T^{n}}$

$\sf{\underline{Let\:us\:find}}$:

The dimensions of current in known units.

$\sf{\underline{So}}$:

$\boxed{\sf{Force = Charge \times Electric\:field}}$

$\sf{\underline{Which\:is\:same\:as}}$:

$\boxed{\sf{Force = \frac{Current \times Time \times Voltage}{Distance}}}$

$\sf{\underline{We\:know\:that}}$:

$\sf{Current = \frac{Force \times Distance}{(Time \times Voltage)}}$

$\sf{Current = \frac{Force \times Distance}{(Time \times Current \times Resistance)}}$

$\implies$ $\sf{I = \frac{M L T^{-2} \times L}{(T \times I \times R)}}$

$\implies$ $\sf{I = M L^{2} T^{-3} I^{-1} R^{-1}}$

$\implies$ $\sf{I^{2}=ML^{2}T^{-3}R^{-1}}$ 

$\sf{\underline{Note}}$: Check attachment (No. 1)

$\boxed{\sf{Energy=I^{k}R^{m}T^{n}}}$

$\sf{\underline{Note}}$: Check attachment (No. 2)

Equating the powers of unit:

$\implies$ k = 2

$\implies$ m = 1

$\implies$ n = 1

$\sf{\underline{Hence}}$:

$\huge\boxed{\sf{H = I^{2} R T}}$