write a short note on token currency of muhamud bin tughluq
Answers
Answer:-
\mathsf{a = \dfrac{g}{7}}a=
7
g
\mathsf{T_2 = \dfrac{6g}{7}}T
2
=
7
6g
Option → C
Given :-
\begin{lgathered}m_3 = 1 kg \\ m_2 = 3kg \\ m_1 = 3 kg\end{lgathered}
m
3
=1kg
m
2
=3kg
m
1
=3kg
Where,
m_3 = a, m_2 = b, m_1 = cm
3
=a,m
2
=b,m
1
=c
To find :-
The acceleration of the block of mass 1 kg and Tension force between A and B.
Solution:-
Let T_1T
1
be the Tension force between Block B and Block C
and T_2T
2
be the tension force between B and A.
Now,
Consider Block C as system.
The force acting on it :-
Mg force downward.
Tension force upward.
Acceleration is upward.
\mathsf{m_1 g - T_1 = m_1 a}m
1
g−T
1
=m
1
a
\mathsf{T_1 = m_1g - m_1 a}----1)T
1
=m
1
g−m
1
a−−−−1)
For Block B.
\mathsf{m_2g + T_2 - T_1 = m_2a}----2)m
2
g+T
2
−T
1
=m
2
a−−−−2)
Take Block A as system.
\mathsf{ m_3g - T_2 = m_3a}m
3
g−T
2
=m
3
a
\mathsf{T_2 = m_3g - m_3 a}-------3)T
2
=m
3
g−m
3
a−−−−−−−3)
Put the value ofT_1, T_2T
1
,T
2
in equation 2.
→\mathsf{m_2g + m_3g - m_3a -(m_1g -m_1a) = m_2a}m
2
g+m
3
g−m
3
a−(m
1
g−m
1
a)=m
2
a
→\mathsf{m_2g + m_3g - m_3a - m_1g - m_1a = m_2a }m
2
g+m
3
g−m
3
a−m
1
g−m
1
a=m
2
a
→\mathsf{m_2g + m_3g - m_1 g = m_2a + m_3a + m_1a}m
2
g+m
3
g−m
1
g=m
2
a+m
3
a+m
1
a
Put the masses of each one.
→\mathsf{3g + g - 3g = 3a + a + 3a }3g+g−3g=3a+a+3a
→\mathsf{ g = 7a }g=7a
→\mathsf{a = \dfrac{g}{7}}a=
7
g
put the value of a in eq. 3
→\mathsf{T_2 = m_3g - m_3a}T
2
=m
3
g−m
3
a
→\mathsf{T_2 = g - a}T
2
=g−a
→\mathsf{T_2 = g - \dfrac{g}{7}}T
2
=g−
7
g
→\mathsf{T_2 = \dfrac{7g-g}{7}}T
2
=
7
7g−g
→\mathsf{T_2 = \dfrac{6g}{7}}T
2
=
7
6g
hence,
The acceleration is \dfrac{g}{7}
7
g
and Tension force between A and B is \dfrac{6g}{7}
7
6g