Question

Two bodies of masses 0.03 kg, and 0.04 kg,
are tied to the ends of a massless string. This
string passes over the frictionless pulley. The
tension in the string is​

Answer 1

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{Tension=\frac{2.4}{7}\:N}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline \bold{Given :}} \\ \tt: \implies Mass \: of \: body (m_{1}) = 0.03 \: kg \\ \\ \tt: \implies Mass \: of \: body (m_{2}) = 0.04 \: kg \\ \\ \red{\underline \bold{To \: Find :}} \\ \tt: \implies Tension \: in \: the \: string(T) = ?$

• According to given question :

$\tt \circ \: Let \: assume \: block \: of \: mass \: m_{2} \: will \\ \tt \: \: \: go \: down \: with \: acceleration \: (a) \: and \\ \tt \: \: \: m_{1} \: will \: go \: up \: with \:same \: acceleration \: (a) \\ \\ \bold{As \: we \: know \: that} \\ \tt: \implies T - m_{1}g = m_{1}a \\ \\ \tt \circ \: Acceleration \: due \: to \: gravity = 10 \: {m/s}^{2} \\ \\ \tt: \implies T- 0.03 \times 10 = 0.03 \times a \\ \\ \tt: \implies T - 0.3 = 0.03a - - - - - (1) \\ \\ \bold{Similarly:} \\ \tt: \implies m_{2}g - T= m_{2}a \\ \\ \tt: \implies 0.04 \times 10 - T = 0.04 \times a$

$\tt: \implies 0.4 -T = 0.04a - - - - - (2)\\ \\ \text{Adding\: (1) \:and \: (2)} \\ \\ \tt: \implies 0.4 - 0.3 = 0.03a + 0.04a \\ \\ \tt: \implies 0.1 = (0.03 + 0.04)a \\ \\ \tt: \implies a = \frac{0.1}{0.07} \\ \\ \tt: \implies a = \frac{10}{7} \: m /{s}^{2} - - - - - (3) \\ \\ \text{Putting \: value \: of \: (3) \: in \: (1)} \\ \\ \tt: \implies T - 0.3 = 0.03 \times \frac{10}{7} \\ \\ \tt: \implies T = \frac{0.3}{7} + 0.3 \\ \\ \tt: \implies T= \frac{0.3 + 2.1}{7} \\ \\ \green{ \tt: \implies T = \frac{2.4}{7} \: N} \\ \\ \green{\tt \therefore Tension \: in \: string \: is \: \frac{2.4}{7} \: N}$

Answer 2

Answer

Tension= $7</h3><p>2.4</p><p> </p><p> N$

$\ \ \\ {\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\begin{gathered}\ \\ {\underline \bold{Given :}} \\ \tt: \implies Mass \: of \: body (m_{1}) = 0.03 \: kg \\ \\ \tt: \implies Mass \: of \: body (m_{2}) = 0.04 \: kg \\ \\ \blue{\underline \bold{To \: Find :}} \\ \tt: \implies Tension \: in \: the \: string(T) = ?\end{gathered}$

$\begin{gathered}\tt \circ \: Let \: assume \: block \: of \: mass \: m_{2} \: will \\ \tt \: \: \: go \: down \: with \: acceleration \: (a) \: and \\ \tt \: \: \: m_{1} \: will \: go \: up \: with \:same \: acceleration \: (a) \\ \\ \bold{As \: we \: know \: that} \\ \tt: \implies T - m_{1}g = m_{1}a \\ \\ \tt \circ \: Acceleration \: due \: to \: gravity = 10 \: {m/s}^{2} \\ \\ \tt: \implies T- 0.03 \times 10 = 0.03 \times a \\ \\ \tt: \implies T - 0.3 = 0.03a - - - - - (1) \\ \\ \bold{Similarly:} \\ \tt: \implies m_{2}g - T= m_{2}a \\ \\ \tt: \implies 0.04 \times 10 - T = 0.04 \times a\end{gathered}$

$</p><p>\begin{gathered}\tt: \implies 0.4 -T = 0.04a - - - - - (2)\\ \\ \text{Adding\: (1) \:and \: (2)} \\ \\ \tt: \implies 0.4 - 0.3 = 0.03a + 0.04a \\ \\ \tt: \implies 0.1 = (0.03 + 0.04)a \\ \\ \tt: \implies a = \frac{0.1}{0.07} \\ \\ \tt: \implies a = \frac{10}{7} \: m /{s}^{2} - - - - - (3) \\ \\ \text{Putting \: value \: of \: (3) \: in \: (1)} \\ \\ \tt: \implies T - 0.3 = 0.03 \times \frac{10}{7} \\ \\ \tt: \implies T = \frac{0.3}{7} + 0.3 \\ \\ \tt: \implies T= \frac{0.3 + 2.1}{7} \\ \\ \pink{ \tt: \implies T = \frac{2.4}{7} \: N} \\ \\ \orange{\tt \therefore Tension \: in \: string \: is \: \frac{2.4}{7} \: N}\end{gathered}$