In terms of absolute error, T = (2.62\(\pm \) 0.11)s.
In terms of percentage error, T = (2.62\(\pm \) 4%)s.

Five

The dimensions of LHS
= [M][LT^-2] = [ML²T^-2]
The dimensions of RHS = [M][LT^-1] = [ML²T^-2]
The dimensions of LHS and RHS are same and hence the consistency is verified.

Two simple harmonic motions of equal amplitudes (A) and frequencies ω and 2ω and initial phase difference of π/2 are represented by
x = A sin ωt     ...(i)
\(y=A\ sin\left({2ω+{\pi\over 2}}\right)=A\ cos\ 2 \omega t\)       ...(ii)
Since cos 2 ωt (1 - 2 sin² ωt)
∴ y = A[1 - 2 sin² ωt]      ...(iii)
From eqn. (i) 
\(sin^2\ ωt={x^2\over A^2}\)
\(∴\ \ y=A\left[1-{2x^2\over A^2}\right]=A-{2x^2\over A}\)
\(⇒\ \ {2x^2\over A}+y-A=0\)
or \(x^2+{Ay\over 2P}-{A^2\over 2}=0\)
which is the equation of a parabola. Hence the resultant path followed by the particle is parabolic.

Given, A = 6 cm = 0.06m
v = 200 Hz
k = \(\frac { 2\pi }{ \lambda } =\frac { 2\pi }{ v/v } \)
k = \(\frac { 2\pi v }{ v } =\frac { 2\pi \times 200 }{ 400 } =\pi { m }^{ -1 }\)
 \(\omega \) = \(2\pi v=2\pi \times 200=400\pi rad/s\)
The standard equation of the progressive wave is
y(x,t) = Asin(kx -\(\omega t\))
Put the values to get the equation
y(x,t) = 0.06 sin\((\pi x-400\pi t)m\) 

Given, \({ m }_{ 1 }=m,{ \quad v }_{ 1 }=v,\quad { V }_{ 1 }=V,\quad { P }_{ 1 }=?\)
 \({ m }_{ 2 }=\frac { m }{ 2 } ,\ { v }_{ 2 }=2v,\ { V }_{ 2 }={ V }_{ 1 }=V,\quad { p }_{ 2 }=?\)
\( { P }_{ 1 }=\frac { 1 }{ 3 } \frac { { m }_{ 1 } }{ { V }_{ 1 } } { v }_{ 1 }^{ 2 }\ and\ { p }_{ 2 }=\frac { 1 }{ 3 } \frac { { m }_{ 2 } }{ { V }_{ 2 } } { v }_{ 2 }^{ 2 }\)
\(\\ \therefore \frac { { P }_{ 1 } }{ { p }_{ 2 } } =\frac { { m }_{ 1 } }{ { m }_{ 2 } } \times \frac { { V }_{ 1 } }{ { V }_{ 2 } } \times \frac { { v }_{ 1 }^{ 2 } }{ { v }_{ 2 }^{ 2 } } \)
\(\\ =\frac { m }{ m/2 } \times \frac { V }{ V } \times \left( \frac { v }{ 2v } \right) ^{ 2 }\)
\(=2\times \frac { 1 }{ 4 } =\frac { 1 }{ 2 } \)

(i) Aman loves nature and his helpful boy. He also has presence of mind as he thought of an excellent idea to help cartman.
(ii) \({ L }_{ 1 }=5.231m,{ L }_{ 2 }=5.243m,{ T }_{ 1 }=27 ^{0}C,{ T }_{ 2 }=?\)
As we know 
\(\therefore { T }_{ 2 }-{ T }_{ 1 }=\frac { { L }_{ 2 }-{ L }_{ 1 } }{ { L }_{ 1 }\times \alpha } \)
\(\\ { T }_{ 2 }=\frac { { L }_{ 2 }-{ L }_{ 1 } }{ { L }_{ 1 }\alpha } \)
\({ T }_{ 2 }=\frac { 5.243-5.231 }{ 5.231\times 1.2\times { 10 }^{ -5 } } +27\)
\(\\ =218 \ ^{0} C\)
(iii) Linear expansion of solids i.e increase in length of solid on heating.

Total cross-sectional area of the femurs is A = 2 × 10 cm² = 20 × 10^–4 m² . The force acting on them is F = 40 kg wt = 400 N (taking g = 10 m s^–2). This force is acting vertically down and hence, normally on the femurs. Thus, the average pressure is
\({ P }_{ av }=\frac { F }{ A } =\frac { 400 }{ 20\times { 10 }^{ -4 } } \)