Question

Hola
Happy diwali
Please answer this!!​

Answer 1

Explanation:

$hello... \\ given \: that = > Area \: = 1 {cm}^{2} \\ length \: of \: smaller \: arm \: = 50 \: cm \: \\ length \: of \: longer \: arm = 100 \: cm \\ pressure = > 80cm \: of \: hg \\ mercury \: rises \: in \: shorter \: tube \: by \: 10 \: cm \\ \\ \\ suppose \: p1 \: and \: p2 \: are \: the \: pressures \\ \: in \: (cm \: of \: hg) \: and \: let \: l1 \: be \: rise \: in \: second \: arm \: \\ that \: we \: have \: to \: find.. \\ \\ \\ we \: know \: that \: temperature \: is \: constant \: so \: we \: \\ can \: directly \: write \: = > \\ PV = k(constant) \\ \\ but \: for \: p1 \: and \: p2 \: pressures = > \\ \\now \: for \: shorter \: arm \\ \\ PV = p1(50 - 10) \\ (we \: minused \: 10 \: because \: it \: had \: \\ risen \: and \: the \: volume \: is \: minimised \\ \\ 80 \times 50 \times 1 = p1(40) \\ p1 = 100 \: cm \\ \\ now \: for \: longer \: arm \\ 80 \times 100 \times 1 = p2(100 - l1) \\ (used \: same \: logic \: as \: above) \\ \\ now \\ p1 = p2(l1 - 10) \\ (change \: in \: mercury \: levels) \\ p2 = 110 - l1 \\ \\ substitute \: p2 = 110 - l1 \\ (110 - l1)(100 - l1) = 8000 \\ \\ {l1}^{2} - 210 \: l1 + 3000 = 0 \\ from \: here \: you \: will \: get \: l1 \: as \: 15.5 \: cm \\ \\ yah \: we \: got \: answer \: \\ so \: our \: required \: length \: is \: 15.5 \: cm$

Answer 2

$\begin{lgathered}hello... \\ given \: that = > Area \: = 1 {cm}^{2} \\ length \: of \: smaller \: arm \: = 50 \: cm \: \\ length \: of \: longer \: arm = 100 \: cm \\ pressure = > 80cm \: of \: hg \\ mercury \: rises \: in \: shorter \: tube \: by \: 10 \: cm \\ \\ \\ suppose \: p1 \: and \: p2 \: are \: the \: pressures \\ \: in \: (cm \: of \: hg) \: and \: let \: l1 \: be \: rise \: in \: second \: arm \: \\ that \: we \: have \: to \: find.. \\ \\ \\ we \: know \: that \: temperature \: is \: constant \: so \: we \: \\ can \: directly \: write \: = > \\ PV = k(constant) \\ \\ but \: for \: p1 \: and \: p2 \: pressures = > \\ \\now \: for \: shorter \: arm \\ \\ PV = p1(50 - 10) \\ (we \: minused \: 10 \: because \: it \: had \: \\ risen \: and \: the \: volume \: is \: minimised \\ \\ 80 \times 50 \times 1 = p1(40) \\ p1 = 100 \: cm \\ \\ now \: for \: longer \: arm \\ 80 \times 100 \times 1 = p2(100 - l1) \\ (used \: same \: logic \: as \: above) \\ \\ now \\ p1 = p2(l1 - 10) \\ (change \: in \: mercury \: levels) \\ p2 = 110 - l1 \\ \\ substitute \: p2 = 110 - l1 \\ (110 - l1)(100 - l1) = 8000 \\ \\ {l1}^{2} - 210 \: l1 + 3000 = 0 \\ from \: here \: you \: will \: get \: l1 \: as \: 15.5 \: cm \\ \\ yah \: we \: got \: answer \: \\ so \: our \: required \: length \: is \: 15.5 \: cm\end{lgathered}$