RIGHTARROW V_x = \frac{{dx}}{{dt}} = {u_0}\left( {1 + \frac{{3x}}{L}} \right)\)SEPARATION of variable\(\Rightarrow \frac{{dx}}{{{u_0}\left( {1 + \frac{{3x}}{L}} \right)}} = dt\)Integrating the equation form 0 to T (Required TIME)\(\mathop \smallint \limits_0^T dt = \mathop \smallint \limits_0^L \frac{{dx}}{{{u_0}\left( {1 + \frac{{3x}}{L}} \right)}} = \frac{1}{{{u_0}}}\frac{L}{3} \cdot \left[ {\ln \left( {1 + \frac{{3x}}{L}} \right)} \right]_0^L\)\(T = \frac{1}{{{u_0}}}\left[ {\ln \left( {1 + \frac{{3x}}{L}} \right)} \right]_0^L \cdot \left( {\frac{L}{3}} \right)\)\(= \frac{L}{{3{u_0}}}{\rm{ln}}\left[ {1 + 3} \right]\)\(T = \frac{L}{{3{u_0}}}\ln \left[ 4 \right]\)

"> RIGHTARROW V_x = \frac{{dx}}{{dt}} = {u_0}\left( {1 + \frac{{3x}}{L}} \right)\)SEPARATION of variable\(\Rightarrow \frac{{dx}}{{{u_0}\left( {1 + \frac{{3x}}{L}} \right)}} = dt\)Integrating the equation form 0 to T (Required TIME)\(\mathop \smallint \limits_0^T dt = \mathop \smallint \limits_0^L \frac{{dx}}{{{u_0}\left( {1 + \frac{{3x}}{L}} \right)}} = \frac{1}{{{u_0}}}\frac{L}{3} \cdot \left[ {\ln \left( {1 + \frac{{3x}}{L}} \right)} \right]_0^L\)\(T = \frac{1}{{{u_0}}}\left[ {\ln \left( {1 + \frac{{3x}}{L}} \right)} \right]_0^L \cdot \left( {\frac{L}{3}} \right)\)\(= \frac{L}{{3{u_0}}}{\rm{ln}}\left[ {1 + 3} \right]\)\(T = \frac{L}{{3{u_0}}}\ln \left[ 4 \right]\)

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In a steady flow through a nozzle, the flow velocity on the nozzle axis is given by \(v\; = \;u_o\left( {1 + \frac{{3{\rm{x}}}}{L}} \right){\rm{}},\) where x is the distance along the axis of the nozzle from its inlet plane and L is the length of the nozzle. The time required for a fluid particle on the axis to travel from the inlet to the exit plane of the nozzle is

BITSAT Kinematics in BITSAT 9 months ago

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\(V = {u_0}\left( {1 + \frac{{3x}}{L}} \right)i\)\(\RIGHTARROW V_x = \frac{{dx}}{{dt}} = {u_0}\left( {1 + \frac{{3x}}{L}} \right)\)SEPARATION of variable\(\Rightarrow \frac{{dx}}{{{u_0}\left( {1 + \frac{{3x}}{L}} \right)}} = dt\)Integrating the equation form 0 to T (Required TIME)\(\mathop \smallint \limits_0^T dt = \mathop \smallint \limits_0^L \frac{{dx}}{{{u_0}\left( {1 + \frac{{3x}}{L}} \right)}} = \frac{1}{{{u_0}}}\frac{L}{3} \cdot \left[ {\ln \left( {1 + \frac{{3x}}{L}} \right)} \right]_0^L\)\(T = \frac{1}{{{u_0}}}\left[ {\ln \left( {1 + \frac{{3x}}{L}} \right)} \right]_0^L \cdot \left( {\frac{L}{3}} \right)\)\(= \frac{L}{{3{u_0}}}{\rm{ln}}\left[ {1 + 3} \right]\)\(T = \frac{L}{{3{u_0}}}\ln \left[ 4 \right]\)

Posted on 18 Nov 2024, this text provides information on BITSAT related to Kinematics in BITSAT. Please note that while accuracy is prioritized, the data presented might not be entirely correct or up-to-date. This information is offered for general knowledge and informational purposes only, and should not be considered as a substitute for professional advice.

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