Question

who will answer this question i ll mark him/ her brainliest answer​

Answer 1

Answer:

c is correct i think

Explanation:

Answer 2

Answer:

• C. 2 : 1

Explanation:

• Resistance α $\sf \frac{L}{D²}$ _____(1)

Where,

L = Length

D = Diameter

According To The Question,

The ratio Of resistance of P To that Of Q is,

Hence,

• $\sf \frac{R_P}{R_Q}$ _____(2)

• $\sf R_Q = \frac{L_Q}{{D_Q}^2}$

• $\sf R_P = \frac{L_P}{{D_P}^2}$

As Given Condition,

---------------

$\sf L_P = \frac{L_Q}{2}$ _____(3)

---------------

$\sf D_P = \frac{D_Q}{2}$ ______(4)

Where, D = Diameter

---------------

According Equation_(1)

So,

• $\sf \frac{R_P}{R_Q} = \frac{ \frac{L_P}{{D_P}^2}}{ \frac{L_Q}{{D_Q}^2}}$

• $\sf \frac{R_P}{R_Q} = \frac{L_P}{{D_P}^2} \times \frac{{D_Q}^2}{L_Q}$

• $\sf \frac{R_P}{R_Q} = \frac{L_P{D_P}^2} {{D_Q}^2L_Q}$

Now, According To Equation (3) and (4)

• $\sf \frac{R_P}{R_Q} = \frac{ (\frac{L_Q}{2}){D_P}^2} { (\frac{D_P}{2} )^{2} L_Q}$

• $\sf \frac{R_P}{R_Q} = \frac{ L_Q{D_P}^2} {( \frac{{D_P }^{2} }{2} ) L_Q}$

• $\sf \frac{R_P}{R_Q} = \frac{ 2L_Q{D_P}^2} {( {D_P }^{2}) L_Q}$

• $\sf \frac{R_P}{R_Q} = \frac{ 2 \: \cancel{L_Q}{D_P}^2} {( {D_P }^{2}) \cancel{L_Q}}$

• $\sf \frac{R_P}{R_Q} = \frac{ 2 \: \cancel{ {D_P}^2}} {( \cancel{ {D_P }^{2}) }}$

• $\sf \frac{R_P}{R_Q} = \frac{2}{1}$

Hence,

• $\sf \large {\boxed{\frac{R_P}{R_Q} = 2 : 1}}$

Option (C) is right.

I hope it helps you...