21for the thermo question it should be (b) and (d) as efficiency is 1-(T2/T1). so efficiency comes out to be 0.33 and since input is 1 kilocalorie which is 4200 joules. so from equation w=eff*heat input the work done comes to be 1400 j.
3okie [1]
3 .
see initial pressure was P0 .........
P0h = P(h-x) .....
P0h/(h-x) = P ..........
SO BALENCING PRESSURES IN EQUILIBRIUM ...
P + Ïgx = Ïgh + P0 .....
P0[h/(h-x)-1] =Ïg(h-x)...
P0 = Ïgh [given] .......
h[x/(h-x)] = h-x...
hx = (h-x)2 .....
may hav made calculation mistakes but im sure of my method
similar kinda prob has been done before........
also im assumin the temp. is constant
3machan wat u have done i cant understand da.can u explain.neways u have given rong ans da.
y P(h-x)=P0h
3UR OPTIONS ARE ALSO RONG DA .......
THE OPTIONS UVE MARKED A.......
h(h-x) = h2 ...
h = h-x ....
x =0 .....
clearly its rong!!!!!!!![3] ........
luks like printin error ... watz the ans. in the book!!!!!!!!!!!!!!!
3no it is not the ans i rote it is one of the rong ans.ne ways the ans is b
3sankara open out the brackets in my answer and answer b ull get same thing so my answer is rite [1]
62both of these are the same answers....
b and what iitimcomin have got!
3(2h-x)(h-x) = h2 ....
2h2 - 2hx - hx + x2 = h2 ......
h2 - 3hx + x2 = 0 .......
now my answer is
hx = (h-x)2 .....
hx = h2 + x2 - 2hx ........
h2 - 3hx + x2 = 0........
which is same as the correct answer b .......
so my answer is rite[1]!!!!!!..just didnt simplify it!!!!!
3W/Qin = neta .........
1- Qout/Qin = neta ........
Qin/Tsource = Qout/Tsink
1- Tsink/Tsource = neta ....
T is in kelvin ..... no u can calculate it i think!!!!!!!
and the amount of heat performed per calorie of heat input
= neta * 4.14 * 100 ....(if i got conversions rite [3]).....
[1]
3sankara ...... mechanics ellai na thermo lende question poode da........... please[2]
