You visited us 1 times! Enjoying our articles? Unlock Full Access!

Consistency of Linear System of Equations

The length of...

Question

The length of wire when $M_{1}$ is hung from it is $l_{1}$
and is $l_{2}$ with both $M_{1}$ and $M_{2}$ hanging. The natural length of wire is

\frac{M_{1}}{M_{2}} (l_{1} - l_{2}) + l_{2}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

\frac{M_{2} l_{1} - M_{1} l_{2}}{M_{1} + M_{2}}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{l_{1} + l_{2}}{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\sqrt{l_{1}, l_{2}}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $\frac{M_{1}}{M_{2}} (l_{1} - l_{2}) + l_{2}$
$\begin{matrix} Mass = M_{1} length = l_{1} mass = M_{1} + M_{2} length = l_{2} Let the natural length be x . using \frac{F}{A} = (\frac{Δ L}{L}) Y, \end{matrix}$

(1) $\frac{M_{1} g}{A} = Y \frac{l_{1} - x}{x}$

(2) $\frac{(M_{1} + M_{2}) g}{A} = Y \frac{l_{2} - x}{x}$

$solving (1 \div 2) we obtain,$

$x = \frac{M_{1}}{M_{2}} (l_{1} - l_{2}) + l_{2}$

Suggest Corrections

Similar questions

Q. The length of wire, when

M_{1}

is hung from it is

l_{1}

and is

l_{2}

with both

M_{1}

and

M_{2}

hanging. The natural length of wire is

If two lines have direction cosines

$l_{1}, m_{1}, n_{1}$ and $l_{2}, m_{2}, n_{2}$

then the condition for these lines being perpendicular to each other would be -

(l_{1}, m_{1}, n_{1})

and

(l_{2}, m_{2}, n_{2})

are d.c's two lines. Then
The lines are parallel is

\frac{l_{1}}{l_{2}} = \frac{m_{1}}{m_{2}} = \frac{n_{1}}{n_{2}}

Let L₁ and L₂ be two non-vertical lines with slopes m₁ and m₂ respectively. If α₁ and α₂ are the inclinations of lines L₁ and L₂, respectively. the acute angle (say θ) between lines L1 and L2 , is given by

Q. A wire of length

L

with Young's modulus

Y

is hanging from a fixed support. The length of wire becomes

L_{1}

when mass

M_{1}

is suspended, and it becomes

L_{2}

when mass

M_{2}

is suspended at its free end. Then, the original length

L

is equal to -

Join BYJU'S Learning Program

The length of wire when M1 is hung from it is l1 and is l2 with both M1 and M2 hanging. The natural length of wire is

The correct option is A M1M2(l1−l2)+l2 Mass =M1 length =l1 mass =M1+M2 length =l2 Let the natural length be x . using FA=(ΔLL)Y , (1)M1gA=Yl1−xx(2)(M1+M2)gA=Yl2−xxsolving (1÷2)we obtain,x=M1M2(l1−l2)+l2

The length of wire when $M_{1}$ is hung from it is $l_{1}$
and is $l_{2}$ with both $M_{1}$ and $M_{2}$ hanging. The natural length of wire is