MCQ
A surveyor's $30$-$m$ steel tape is correct at some temperutre. On a hot day the tape has expanded to $30.01$ $m$. On that day, the tape indicates a distance of $15.52$ $m$ between two points. The true distance between these points is :-
  • A
    $15.515 $ $m$
  • B
    $15.520$ $ m$
  • $15.525$ $m$
  • D
    $15$

Answer

Correct option: C.
$15.525$ $m$
c
$30 (1 + \alpha \theta ) = 30.01$
$15.52 (1 + \alpha \theta ) = x$
$x = 15.52 × \frac{30.01}{30}$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Similar questions

The shear strain is possible in .............
A couple produces
Two particles $A$ and $B$ are moving in a straight line with the same speed. Which of the following statement (s) is/ are correct for the relative motion of the two particles?
When an ideal gas is compressed adiabatically, its temperature rises the molecules on the average have more kinetic energy than before. The kinetic energy increases,
A force of $750 \,N$ is applied to a block of mass $102\, kg$ to prevent it from sliding on a plane with an inclination angle $30°$ with the horizontal. If the coefficients of static friction and kinetic friction between the block and the plane are $0.4 $ and $0.3$ respectively, then the frictional force acting on the block is...... $N$
Which of the following statements is correct?
In the figure given below. a block of mass $M =490\,g$ placed on a frictionless table is connected with two springs having same spring constant $\left( K =2 N m ^{-1}\right)$. If the block is horizontally displaced through ' $X$ 'm then the number of complete oscillations it will make in $14 \pi$ seconds will be $.........$
Two springs have their force constant as $k_1$ and $k_2 (k_1 > k_2)$. when they are  stretched by the same force
Two bodies of masses $m$ and $4 \,m$ are moving with equal $K.E.$ The ratio of their linear momentums is
A particle executes simple harmonic motion with an amplitude of $4 \mathrm{~cm}$. At the mean position, velocity of the particle is $10 \mathrm{~cm} / \mathrm{s}$. The distance of the particle from the mean position when its speed becomes $5 \mathrm{~cm} / \mathrm{s}$ is $\sqrt{\alpha} \mathrm{cm}$, where $\alpha=$____________.