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CIE AS LEVEL PHYSICS MULTIPLE CHOICE DOUTS

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nid404:
Rest when I get back home.

S.M.A.T:
4)G.P.E=mgh=(1.3*10^9)*9.81*2=2.55*10^10 J
Therefore output power for one day is

Power=Energy/time=(2.55*10^10)/(1*24*60*60)=2.95*10^5 which is approximately 300kW.

S.M.A.T:
5)Initially the emf of the cell(in the bottom circuit) is equal to the potential difference in wire XN so galvanometer shows no deflection.But when the resistance of variable resistor is increased the p.d across variable resistor increases so the p.d across XY decreases.(because total p.d in the(top) circuit remain the same).Now the potential difference across XN is less than the emf of the cell so the movable contact must be move towards the Y to increase the p.d across XN.so when the p.d become equal to emf of the cell the galvanometer again shows zero deflection.

Summarizing,

Initial
Emf of the cell in the bottom circuit=pd across XN.
After increasing resistance of variable resistor,the p.d across XY decreases.
Now p.d across XN is less than before,so the movable contact must be move towards the Y to increase the p.d across XN.

S.M.A.T:
7)The answer would be c because in the region between two parallel charge plate,the equipotential lines are equally spaced.So as the distance 'd' increase voltage increases proportionally so 'E' remain constant as E=V/d.In the region between two parallel charge plate the electric field strength is always constant

nid404:
Thanks +rep

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