Voltmeter Reading with Ideal Diode and Fully Charged Capacitor (10V & 5V Connected)

The diode is an ideal one. No forward resistance.
The capacitor fully charged to given volt in figure.
Now 10 volt and 5 volt is connected. Wat is voltmeter value

this is the same question you asked before. i would think the capacitor's voltage adds in series with the 10 V cell assuming you meant the direction of your 10 V battery has the positive end on the longer plate. if this is the case, then Vab is 10 + 15 V which is always greater than the 5 V so the diode is never forward-biased, so the current won't flow through the diode or the capacitor. with no current flow, the capacitor will hold its charge, so vab = 25 V, Vdiode (reverse-biased) = 20 V.

Sir, wat u said is correct Vab=25v. Now my doubt is wat would be Vab, if there was no diode, that is short circuited and everything else same as in question. Wat would be voltmeter value?

if you are removing the diode, then the voltage across the capacitance would conform to the two batteries, so instead of the current 15 V, it would be -5 V if we maintain the orientation of the capacitor.

it is the same as if you imagine vb = ground reference and you connect a capacitor across two batteries.

as for vab, since you turned the diode into a short, then you are just measuring the voltage across the 5 V battery. vab = 5 V