2. An object starting from rest travels 120 m in first 2 s and 260 m in next 4 s. Find
its velocity in 6 s after it starts.
Answers
Answer:
Given :
The period of revolution (t) of a planet around the sun depends upon radius (r) of the orbit, mass (m) of the sun and gravitational constant (G), mathematically :
\implies \sf [T] \propto [r]^a [m]^b [G]^c⟹[T]∝[r]
a
[m]
b
[G]
c
Adding k as dimensionaless constant :
\begin{gathered}\implies \sf [T] = k [r]^a [m]^b [G]^c \: \: \: \: \:... (i)\\ \end{gathered}
⟹[T]=k[r]
a
[m]
b
[G]
c
...(i)
• DIMENSIONS:
\begin{gathered}\dag \bf \: Dimensions \: of \: period\:of\: revolution \: (t) = [M^0L^0T^1] \\ \end{gathered}
†Dimensionsofperiodofrevolution(t)=[M
0
L
0
T
1
]
\begin{gathered}\dag \bf \: Dimensions \: of \: radius \: (r) = [M^0L^1T^0] \\ \end{gathered}
†Dimensionsofradius(r)=[M
0
L
1
T
0
]
\begin{gathered}\dag \bf \: Dimensions \: of \: mass \: (m) = [M^1L^0T^0] \\\end{gathered}
†Dimensionsofmass(m)=[M
1
L
0
T
0
]
\begin{gathered}\dag \bf \: Dimensions \: of \: gravitational \:constant \: (G) = [M^{ - 1} L^3T^{ - 2} ] \\\end{gathered}
†Dimensionsofgravitationalconstant(G)=[M
−1
L
3
T
−2
]
\begin{gathered}\implies \sf [M^0L^0T^1] =[M^0L^1T^0] ^a [M^1L^0T^0]^b [M^{ - 1} L^3T^{ - 2} ]^c \\ \end{gathered}
⟹[M
0
L
0
T
1
]=[M
0
L
1
T
0
]
a
[M
1
L
0
T
0
]
b
[M
−1
L
3
T
−2
]
c
\begin{gathered}\implies \sf [M^0L^0T^1] =[M]^{b - c} [L]^{a + 3c} [T]^{ - 2c} \\ \end{gathered}
⟹[M
0
L
0
T
1
]=[M]
b−c
[L]
a+3c
[T]
−2c
• On comparing the powers we have:
\implies b - c = 0⟹b−c=0
\implies b = c \: \: \: ....(ii)⟹b=c....(ii)
\implies a + 3c = 0 \: \: \: ....(iii)⟹a+3c=0....(iii)
\implies - 2c = 1⟹−2c=1
\implies \bf c = \dfrac{ -1 }{2} \: \: \: ....(iv)⟹c=
2
−1
....(iv)
Substituting value of c = -1/2 from equation (iv) to equation (ii) :
\begin{gathered}\implies b = c \\ \end{gathered}
⟹b=c
\implies \bf b = \dfrac{ - 1}{2}⟹b=
2
−1
Also substitue the value of c from equation (iv) to equation (ii)
\implies a + 3c = 0⟹a+3c=0
\implies a + 3 \times \dfrac{ - 1}{2} = 0⟹a+3×
2
−1
=0
\implies a - \dfrac{ 3}{2} = 0⟹a−
2
3
=0
\implies \bf a = \dfrac{ 3}{2}⟹a=
2
3
Now, Substituting the value of a, b and c in the equation (I):
\begin{gathered}\implies \sf [T] = k [r]^{ \frac{3}{2} }[m]^{ \frac{ - 1}{2} } [G]^{ \frac{ - 1}{2} } \\ \end{gathered}
⟹[T]=k[r]
2
3
[m]
2
−1
[G]
2
−1
\begin{gathered}\implies \sf [T]^{2} = k [r]^{3 }[m]^{ - 1 } [G]^{ - 1 } \\ \end{gathered}
⟹[T]
2
=k[r]
3
[m]
−1
[G]
−1
\implies \sf [T]^{2} = k [r]^{3 }⟹[T]
2
=k[r]
3
\implies \sf [T]^{2} \propto [r]^{3 }⟹[T]
2
∝[r]
3
\implies \underline{ \boxed{ \orange{\bf T^{2} \propto r^{3 }}}}⟹
T
2
∝r
3
Hence, Proved!