Chapter Chosen

ચાકગતિ

Book Chosen

ભૌતિક વિજ્ઞાન ધોરણ 11 સેમિસ્ટર 2

Subject Chosen

ભૌતિક વિજ્ઞાન

Book Store

Download books and chapters from book store.
Currently only available for
CBSE Gujarat Board Haryana Board

Previous Year Papers

Download the PDF Question Papers Free for off line practice and view the Solutions online.
Currently only available for
Class 10 Class 12

દ્રઢ પદાર્થની ભ્રમણાક્ષથી 10 cm અંતરે આવેલા કણની કોણીય ઝડપ 12 space r a d space s to the power of negative 1 end exponent છે, તો તે ભ્રમણાક્ષથી 20 cm અંતરે આવેલા કણની કોણીય ઝડપ કેટલી હશે ?

  • r a d space s to the power of negative 1 end exponent

  • 15 r a d space s to the power of negative 1 end exponent

  • 12 r a d space s to the power of negative 1 end exponent

  • 10 r a d space s to the power of negative 1 end exponent


C.

12 r a d space s to the power of negative 1 end exponent

Tips: -

દ્રઢ પદાર્થની વ્યાખ્યા પરથી તેના દરેક કણનો નિયત ભ્રમણાક્ષને અનુલક્ષીને કોણીય વેગ સમાન હોય છે. તેથી 20 cm અંતરે આવેલા કણની કોણીય ઝડપ 12 r a d space s to the power of negative 1 end exponent હશે.


ઘડિયાળના મિનિટ-કાંટાની કોણીય ઝડપ કેટલી ?

  • straight pi over 43200 space rad space straight s to the power of negative 1 end exponent
  • straight pi over 1800 space rad space straight s to the power of negative 1 end exponent
  • straight pi over 6 space rad space straight s to the power of negative 1 end exponent
  • straight pi over 12 space rad space straight s to the power of negative 1 end exponent

B.

straight pi over 1800 space rad space straight s to the power of negative 1 end exponent

Tips: -

મિનિટ-કાંટાને એક ભ્રમણ પૂરું કરતા 1 કલાક = 60 cross times 60 = 3600 s લાગે. એટલે કે 3600 સ માં 2 straight pi space radભ્રમણ કરે છે. 

omega space equals space fraction numerator triangle ϕ over denominator triangle t end fraction space equals space fraction numerator 2 space straight pi over denominator 3600 end fraction space equals space straight pi over 1800 space rad space straight s to the power of negative 1 end exponent


પૃથ્વીની ફરતે ભ્રમણ કરતાં એક કૃત્રિમ ઉપગ્રહનું દળ 500 kg છે. તેનું કોણીય વેગમાન 4 space cross times space 10 to the power of 7 space straight J space straight s હોય, તો તેનો ક્ષેત્રિય વેગ શોધો.

  • 2 space cross times 10 to the power of 4 space straight m squared space straight s to the power of negative 1 end exponent
  • 0

  • 2 space cross times space 10 to the power of 7 space straight m squared space straight s to the power of negative 1 end exponent
  • 4 space cross times space 10 to the power of 4 space straight m squared space straight s to the power of negative 1 end exponent

D.

4 space cross times space 10 to the power of 4 space straight m squared space straight s to the power of negative 1 end exponent

Tips: -

m space equals space 500 space kg space comma space L space equals space 4 space cross times space 10 to the power of 7 space straight J space straight s

ક્ષત્રિય વેગ fraction numerator d A over denominator d t end fraction space equals space 1 half space L over M



                   equals space 1 half space cross times space fraction numerator 4 cross times 10 to the power of 7 over denominator 500 end fraction

equals space 4 cross times 10 to the power of 4 space straight m squared space straight s to the power of negative 1 end exponent

ભ્રમણાથી 10 cm અંતરે આવેલા કણની કોણીય ઝડપ 20 space rad space straight s to the power of negative 1 end exponent છે, તો તેની રેખીય ઝડપ કેટલી ?

  • 1 space cm space straight s to the power of negative 1 end exponent
  • 20 space cm space straight s to the power of negative 1 end exponent
  • 200 space cm space straight s to the power of negative 1 end exponent
  • 400 space cm space straight s to the power of negative 1 end exponent

C.

200 space cm space straight s to the power of negative 1 end exponent

Tips: -

straight nu space equals space rω space equals space left parenthesis 10 right parenthesis left parenthesis 20 right parenthesis space equals space 200 space cm divided by straight s

એક વ્હીલ સ્થિર સ્થિતિમાંથી ગતિ શરૂ કરી 4 s ના અંતે 64 space rad space straight s to the power of negative 1 end exponent જેટલો કોણીય વેગ પ્રાપ્ત કરે છે, તો તેનો અચળ કોણીય પ્રવેગ........હોય.

  • 64 space rad space straight s to the power of negative 2 end exponent
  • 128 space rad space straight s to the power of negative 2 end exponent
  • 16 space rad space straight s to the power of negative 2 end exponent
  • 4 space rad space straight s to the power of negative 2 end exponent

C.

16 space rad space straight s to the power of negative 2 end exponent

Tips: -

omega subscript 0 space equals space 0 comma space omega space equals space 64 space rad space straight s to the power of negative 1 end exponent comma space straight t space equals space 4 space straight s

હવે, alpha space equals space fraction numerator omega space minus space omega subscript 0 over denominator t end fraction space equals space fraction numerator 64 space minus space 0 over denominator 4 end fraction space equals space 16 space rad space straight s to the power of negative 2 end exponent