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1.Units, Dimensions and Measurement
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લીસ્ટ $I$ સાથે લીસ્ટ $II$ યોગ્ય રીતે જોડો.
| લીસ્ટ $I$ (ભૌતિક રાશી) | લીસ્ટ $II$ (પારિમાણિક સૂત્ર) |
| $(A)$ દબાણ પ્રચલન | $(I)$ $\left[ M ^0 L ^2 T ^{-2}\right]$ |
| $(B)$ ઊર્જા-ઘનતા | $(II)$ $\left[ M ^1 L ^{-1} T ^{-2}\right]$ |
| $(C)$ વિદ્યુતક્ષેત્ર | $(III)$ $\left[ M ^1 L ^{-2} T ^{-2}\right]$ |
| $(D)$ ગુપ્ત ઉષ્મા | $(IV)$ $\left[ M ^1 L ^1 T ^{-3} A ^{-1}\right]$ |
A$(A)-(III), (B)-(II), (C)-(I), (D)-(IV)$
B$(A)-(II), (B)-(III), (C)-(IV), (D)-(I)$
C$(A)-(III), (B)-(II), (C)-(IV), (D)-(I)$
D$(A)-(II), (B)-(III), (C)-(I), (D)-(IV)$
(JEE MAIN-2023)
Solution
Pressure gradient$=\frac{ dp }{ dx }=\frac{\left[ ML ^{-1} T ^{-2}\right]}{[ L ]}$
$=\left[ M ^1 L ^{-2} T ^{-2}\right]$
Energy density $=\frac{\text { energy }}{\text { volume }}=\frac{\left[ ML ^2 T ^{-2}\right]}{\left[ L ^3\right]}$
$=\left[ M ^1 L ^{-1} T ^{-2}\right]$
$\text { Electric field }=\frac{\text { Force }}{\text { ch arge }}=\frac{\left\lfloor MLT ^{-2}\right\rfloor}{[ A . T ]}$
$=\left[ M ^1 L ^1 T ^{-3} A ^{-1}\right]$
$\text { Latent heat }=\frac{\text { heat }}{\text { mass }}=\frac{\left[ ML ^2 T ^{-2}\right]}{[ M ]}$
$=\left[ M ^0 L ^2 T ^{-2}\right]$
$=\left[ M ^1 L ^{-2} T ^{-2}\right]$
Energy density $=\frac{\text { energy }}{\text { volume }}=\frac{\left[ ML ^2 T ^{-2}\right]}{\left[ L ^3\right]}$
$=\left[ M ^1 L ^{-1} T ^{-2}\right]$
$\text { Electric field }=\frac{\text { Force }}{\text { ch arge }}=\frac{\left\lfloor MLT ^{-2}\right\rfloor}{[ A . T ]}$
$=\left[ M ^1 L ^1 T ^{-3} A ^{-1}\right]$
$\text { Latent heat }=\frac{\text { heat }}{\text { mass }}=\frac{\left[ ML ^2 T ^{-2}\right]}{[ M ]}$
$=\left[ M ^0 L ^2 T ^{-2}\right]$
Standard 11
Physics
