If the total energy transferred to a surface in time is , then the magnitude of the total momentum delivered to this surface for complete absorption will be

STUDY MATERIALS
COURSES
MORE
SUCCESS STORY
ADMISSION
STORE

Back

JEE ADVANCE

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

JEE MAIN

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

NEET

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

BITSAT

CLASS XI - XII

CLASS XI - XII

CBSE BOARD

CLASS IX

CLASS X

CLASS XI

CLASS XII

CLASS VI

CLASS VII

CLASS VIII

UP BOARD

CLASS IX
CLASS X
CLASS XI
CLASS XII

CLASS IX

CLASS X

CLASS XI

CLASS XII

CUET

CRASH COURSE

CRASH COURSE

Courses

CLASSROOM COURSES
ONLINE COURSES

ONLINE COURSES

Q.
If the total energy transferred to a surface in time $t$ is $6.48 \times 10^{9} J$ , then the magnitude of the total momentum delivered to this surface for complete absorption will be [2024]

1 $2.46 \times 10^{- 3} kgm / s$

2 $2.16 \times 10^{- 3} kgm / s$

3 $1.58 \times 10^{- 3} kgm / s$

4 $4.32 \times 10^{- 3} kgm / s$

Ans.
(2)

Energy transferred in time $' t' = 6.48 \times 10^{5} J$

$Power \Rightarrow P = \frac{6.48 \times 10^{5}}{t}$

For complete absorption

$Pressure = \frac{I}{C} = \frac{Energy}{t . A . C}$

$(Pressure \times A) = \frac{Energy}{time \times C}$

$Force = \frac{Energy}{time \times C}$

$Change in momentum = \frac{Energy}{time \times C}$

So $Momentum delivered = \frac{Energy}{C}$

$= \frac{6.48 \times 10^{5}}{3 \times 10^{8}} = 2.16 \times 10^{- 3} kg m/s$

Solution

Q.
If the total energy transferred to a surface in time $t$ is $6.48 \times 10^{9} J$ , then the magnitude of the total momentum delivered to this surface for complete absorption will be [2024]

1 $2.46 \times 10^{- 3} kgm / s$

2 $2.16 \times 10^{- 3} kgm / s$

3 $1.58 \times 10^{- 3} kgm / s$

4 $4.32 \times 10^{- 3} kgm / s$

Want Us to Email you About Special Offers & Updates?

Solution

Q. If the total energy transferred to a surface in time t is 6.48×109J, then the magnitude of the total momentum delivered to this surface for complete absorption will be [2024]

1 2.46×10-3kgm/s

2 2.16×10-3kgm/s

3 1.58×10-3kgm/s

4 4.32×10-3kgm/s

Ans. (2) Energy transferred in time 't'=6.48×105J Power⇒P=6.48×105t For complete absorption Pressure=IC=Energyt.A.C (Pressure×A)=Energytime×C Force=Energytime×C Change in momentum=Energytime×C So Momentum delivered=EnergyC =6.48×1053×108=2.16×10-3 kg m/s

Want Us to Email you About Special Offers & Updates?

Q.
If the total energy transferred to a surface in time $t$ is $6.48 \times 10^{9} J$ , then the magnitude of the total momentum delivered to this surface for complete absorption will be [2024]

1 $2.46 \times 10^{- 3} kgm / s$

2 $2.16 \times 10^{- 3} kgm / s$

3 $1.58 \times 10^{- 3} kgm / s$

4 $4.32 \times 10^{- 3} kgm / s$