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# Daily Archives: June 16, 2011

## A Problem from power

The power output of a machine that lifts a 600 kg
weight through a height of 20m in 1min is:
@)1.96kW    b)0.98kW

c)12kW      d)3.92kW”

Use the formula power = work /time

$P = \frac{mgh}{t}$

## Fundamental Quantities

“What are the 7 fundamental quantities”

Fundamental Physical quantities are those which can be used to express (almost) all other physical quantities but the fundamental quantities cannot be expressed in terms of each other.

There are 7 Fundamentals quantities identified

1. length (unit metre (m))
2. mass (unit kilogram (kg))
3. time (unit second (s))
4. thermodynamic temperature (unit kelvin (K))
5. intensity of current (unit ampere (A))
6. luminous intensity (Unit candela (Cd)) and
7. amount of substance (unit mole (mol))

## Change in Resistance due to stretching a wire

“A piece of wire is redrawn without any change in volume so that its radius become half the original. Compare the new resistance with the original value.”

When we redraw the wire, the volume remains constant and the resistivity also remains constant. SO, the variables are

1. area of cross section and
2. the length

When radius becomes half, the area of cross section increases in such a way that A1l1 = A2l2. This implies that when length is halved, area of cross section is doubled.

We know that $\large \inline \fn_cs R=\frac{\rho l}{a}$

Therefore, the new resistance becomes (1/4) th the original value.

So, R2:R1 = 1:4

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