MCQ
The unit $J \,P{a^{ - 1}}$ is equivalent to
- ✓${m^3}$
- B$c{m^3}$
- C$d{m^3}$
- DNone of these
Dimension of Joule i.e. work $ = F \times L$ $ = ML{T^{ - 2}} \times L$
$ = \left[ {M{L^2}{T^{ - 2}}} \right]$
$\frac{1}{{Pa}} = \frac{1}{{{\rm{Pressure}}}} = \frac{1}{{\frac{F}{A}}} = \frac{{1 \times A}}{F} = \left[ {ML{T^{ - 1}}} \right]$
So, $J \,P{a^{ - 1}}$ $ = \left[ {M{L^2}{T^2}} \right]$ $ = \left[ {{L^2} \times L} \right]\; = \left[ {{L^3}} \right]$.
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
$[I]$ Only $A, B$ and $C$ react with $1\,M\,HCl$ to give $H_2\,(g)$
$[II]$ When $C$ is added to solutions of the other metal ions, metallic $B$ and $D$ are formed.
$[III]$ Metal $C$ does not reduce $A^{n+}$
| List -$I$ Molecule | List - $II$ Shape |
| $(A)$. $\mathrm{BrF}_5$ | $(I)$. $T$-shape |
| $(B)$. $\mathrm{H}_2 \mathrm{O}$ | $(II)$. See saw |
| $(C)$. $\mathrm{ClF}_3$ | $(III)$. Bent |
| $(D)$. $\mathrm{SF}_4$ | $(IV)$. Square pyramidal |
$C{H_3}COOH\xrightarrow{{LiAl{H_4}}}A\mathop {\xrightarrow{{{H^ + }}}}\limits_{443\,K} B\xrightarrow{{B{r_2}}}C\mathop {\xrightarrow{{alc.}}}\limits_{KOH} D$