Question

In the given figure, O is the centre of the circle, AB and AC are the tangents to the circle at B and C respectively. If ∠BAC = 62°, then the value of ∠BDC is

Answer 1

ANSWER :–

new frequency – \begin{gathered}{ \boxed { \bold { {n}_{1} = \frac{n}{4} }}} \\\end{gathered}

n

1

=

4

n

EXPLANATION :–

GIVEN :–

• The fundamental frequency of a sonometer wire is n.

• The length and diameter of the wire are doubled keeping the tension same.

TO FIND :–

New fundamental frequency = ?

SOLUTION :–

• We know that frequency –

\begin{gathered}\\ { \boxed { \bold { n = \frac{1}{2l} \sqrt{ \frac{T}{m} }}}} \\\end{gathered}

n=

2l

1

m

T

• Mass per unit Length :–

m = [ρΠr²(l)]/l

m = ρΠr²

• So that , new equation –

\begin{gathered}\\ { \boxed { \bold { n = \frac{1}{2(2l)} \sqrt{ \frac{T}{ \rho \pi {(2r)}^{2}}}}}} \\\end{gathered}

n=

2(2l)

1

ρπ(2r)

2

T

• Now According to the question –

Length (l) => 2l and radius (r) => 2r

• New fundamental frequency –

\begin{gathered}\\ { \bold { {n}_{1} = \frac{1}{2(2l)} \sqrt{ \frac{T}{ \rho \pi {(2r)}^{2}} }}} \\\end{gathered}

n

1

=

2(2l)

1

ρπ(2r)

2

T

\begin{gathered}\\ { { \bold { {n}_{1} = \frac{1}{4l} \sqrt{ \frac{T}{4 \rho \pi {r}^{2}} }}}} \\\end{gathered}

n

1

=

4l

1

4ρπr

2

T

\begin{gathered}\\ { { \bold { {n}_{1} = \frac{1}{2(4l)} \sqrt{ \frac{T}{ \rho \pi {r}^{2}} }}}} \\\end{gathered}

n

1

=

2(4l)

1

ρπr

2

T

\begin{gathered}\\ { { \bold { {n}_{1} = \frac{1}{4} {n} }}} \\\end{gathered}

n

1

=

4

1

n

\begin{gathered}\\ { { \bold { {n}_{1} = \frac{n}{4} }}} \\\end{gathered}

n

1

=

4

n