## CBSE Physics Solved Board Question Papers 2008, 2009, 2010 & 2011 – All versions

Download CBSE Physics Solved Board Question Papers 2008, 2009, 2010 & 2011 – All versions (Delhi, Outside Delhi & Foreign)

The question Papers are in PDF format, all question papers with solution in a single file. So, it may take some time for the download to finish.

CBSE Physics Class 12 Board Question Papers 2008 to 2011 all sets (SETS I, II & III) and versions (DELHI, FOREIGN, OUTSIDE DELHI)

## is all physics which we had studied in class 11 and 12 are correct?

Is all physics which we had studied in class 11 and 12 are correct or … ?

I would like to quote just one thing, “Science tells the first word of everything and the last word of nothing”

What we learnt yesterday might have changed today and may change in future. We have to adapt ourselves to the new findings, discoveries and inventions.

Further, What we learn in class XI and XII is up to that level and we when we go to higher levels, we learn things more deeply and many things will be dealt differently.

## The problem of a rock thrown vertically up

A rock is thrown vertically upward from the ground with an initial speed 15m/s.

a. how high does it go
b. how much time is required for the rock to reach its maximum height?
c. what is the rock’s height at t=2.00s?

(Posted by Merhawi)

(a)

u=15m/s

a=-10m/s2

v=0 m/s (at the max height)

S=?

v2-u2=2aS

S=v2-u2/2a

=225/20

=11.25 m

(b) From the above case

using v=u+at

t=v-u/a=1.5s

(c) Use S = ut + 1/2 at2

put t=2s, u = 15m/s, a=-10m/s2

S=30 – 20 = 10 m

(If you use g = 9.8 m/s2 The answers will be slightly different)

## A problem and its solution

Q. A weather balloon is floating at a constant height above earth when it releases a pack of instruments.

a. if the pack hits the ground with a velocity of -73.5 m/s, how far did the pack fall?

b. how long did it take for the pack to fall?
—————————————————————————–
v = -73.5 m/s (- since downwards)
a = -g = -9.8 m/s 2
From the equation of motion
v = u + at
-73.5 = -9.8 x t

t=7.5 s

from the equation of motion s = ut + 1/2 a t2
u= 0 m/s
s = 0.5 x (-9.8)x 7.5 x 7.5
= 275.625 m
= 275.6 m