Question

hey boys, answer my question cbse 10th today's per​

Answer 1

why only boys...

anyways,

Least count of the ammeter:-

diff between two major division/no. of div between them

=>(200-100)/10

=10 mA

Least count of voltmeter:-

(2-1)/10

=0.1V

Resistance R=V/I [Ohm's law]

=0.1/(10/1000)

=10 ohms

Answer 2

$\boxed{\red{\huge{ \mathfrak{heya \: answer - }}}}$

the least count of ammeter

=>there are 10 division between 100ma and 200ma

so the least count is-

$\frac{200 - 100}{10}$

=>10ma

=>10×10-³a

=>10-²

=>0.01

similarity

least count of voltmeter=0.1

and I by observing figure is 250ma(250×10-³a)

and V by observing figure is 2.4V

so the resistance is

V/I

$\frac{250 \times 10 {}^{ - 3} }{2.4}$

$\boxed{9.6}$