JEE PHYSICS MCQ: THERMODYNAMICS

1 Medium

1 of 1

Application +4 | -1

Work done by 0.1 mole of a gas at

{27^ \circ }C

to double its volume at constant pressure is 

\left( {R = 2\;cal\;mo{l^{ - 1}}\,\,^\circ {C^{ - 1}}} \right)

View answer and explanation

Correct option: C

60\,cal

Option C (

2 h

) is correct because the maximum range is achieved at an angle of 45 degrees, and the range is twice the maximum height when the angle for maximum height is considered.

More Questions from THERMODYNAMICS

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The work done in which of the following process is zero? Water is used to cool the radiators of engines in cars because A gas expand

0.25\,\,{m^3}

at constant pressure

{10^3}N/{m^2},

the work done is A thermodynamic process in which temperature

T

of the system remains constant though other variable

P

and

V

may change, is called What is the slope for an isothermal process? For adiabatic processes

\left( {\gamma = \frac{{{C_p}}}{{{C_v}}}} \right)