Question

A labourer is paid Rs.8 per hr for an 8- hr day and 1/2 times that rate for each hour in excess of 8 hrs in a single day. If the labourer received Rs.80 for a single days work, how long did he work on that day?

Answer 1

Answer:

Money paid to the labourer for initial 8 hrs of work=

8×8=₹64.So he receives ₹16 for working extra .As the labourer receives ₹12 per extra one hour. So let's say the number of hours the labourer works be x.=> 12x=16=> x=16/12=4/3 hours=1 hour 20 minutes

Answer 2

$\red{ \qquad \underline{ \pmb{{ \mathbb{ \maltese  \:  COMPLETE \: \: QUESTION \:   \maltese }}}}}$

1)A particle of mass m is thrown upwards from the surface of the earth, with a velocity u. The mass and the radius of the earth are, respectively, M and R. G is gravitational constant and g is acceleration due to gravity on the surface of the earth. The minimum value of u so that the particle does not return back to earth, is?

$\Large\green{\qquad\underline{\pmb{{ \mathbb { \maltese  \:  GIVEN \:  \maltese }}}}}$

1)A particle of mass m is thrown upwards from the surface of the earth, with a velocity u.

2)The mass and the radius of the earth are, respectively, M and R.

3)G is gravitational constant and g is acceleration due to gravity on the surface of the earth.

$\huge\boxed{\fcolorbox{pink}{ink}{TO FIND:}}$

Minimum value of u so that the particle does not return back to earth =?

$\purple{\qquad \qquad \underline{ \pmb{{ \mathbb{ \maltese  \:  REQUIRED \: \: INFO \:  \maltese }}}}}$

$GM=gR²$

$\Large \orange{\qquad \underline{ \pmb{{ \mathbb{ \maltese  \:SOLUTION \:  \maltese }}}}}$

$\large \mathfrak{ \text{W}e \: \text{K}now }$

$V \: e= \sqrt{2gR} = \sqrt{ 2\frac{GM}{ {R}^{2} } }R$

$= \sqrt{ \frac{2GM}{R} }$

• We can also find this question answer by this method also

$\huge\boxed{\fcolorbox{red}{ink}{SOLUTION:}}$

$\small\fbox\red{✯According to law of conservation of Mechanical energy✯}$

$\frac{1}{2}m {u}^{2} - \frac{GMm}{R} = 0$

$OR$

${u}^{2} = \frac{2GM}{R}$

$u = \sqrt{\frac{2GM}{R}} = \sqrt{2gR}$

$(g = \frac{GM}{ {R}^{2} } )$

$= \sqrt{ \frac{2GM}{R} }$

$\huge\fbox\pink{✯Answer✯}$

Minimum value of u so that the particle does not return back to earth =

$= \sqrt{ \frac{2GM}{R} }$

$\huge\boxed{\dag\sf\red{Thanks}\dag}$