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Byju's Answer

Standard XII

Mathematics

nth Term of A.P

Three blocks ...

Question

Three blocks are connected as shown in figure and placed on a horizontal frictionless surface if $m_{1} = 2 k g, m_{2} = 8 k g$ and $m_{3} = 27 k g$ and $T_{3} = 37 N$ then ratio of $T_{2} / T_{1}$ .

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Solution

The correct option is A $5$
Acceleration of system= $\frac{T_{3}}{m_{1} + m_{2} + m_{3}} = 37 / 37 = 1 m / s^{2}$

If we consider $m_{1} m_{2}$ as system then
$T_{2} = (m_{1} + m_{2}) a = 10 \times 1 = 10 N$
Similarly
$T_{1} = m_{1} a = 2 \times 1 = 2 N$

$\frac{T_{2}}{T_{1}} = 5$

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Three blocks are connected as shown in figure and placed on a horizontal frictionless surface if m1=2kg,m2=8kg and m3=27kg and T3=37N then ratio of T2/T1.

The correct option is A 5Acceleration of system=T3m1+m2+m3=37/37=1m/s2If we consider m1m2 as system thenT2=(m1+m2)a=10×1=10NSimilarly T1=m1a=2×1=2NT2T1=5

Three blocks are connected as shown in figure and placed on a horizontal frictionless surface if $m_{1} = 2 k g, m_{2} = 8 k g$ and $m_{3} = 27 k g$ and $T_{3} = 37 N$ then ratio of $T_{2} / T_{1}$ .

The correct option is A $5$
Acceleration of system= $\frac{T_{3}}{m_{1} + m_{2} + m_{3}} = 37 / 37 = 1 m / s^{2}$

If we consider $m_{1} m_{2}$ as system then
$T_{2} = (m_{1} + m_{2}) a = 10 \times 1 = 10 N$
Similarly
$T_{1} = m_{1} a = 2 \times 1 = 2 N$

$\frac{T_{2}}{T_{1}} = 5$