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ભૌતિક વિજ્ઞાન ધોરણ 11 સેમિસ્ટર 2

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ભૌતિક વિજ્ઞાન

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0.100 gનું એક પીંછું  open parentheses negative 0.05 space j with hat on top close parentheses space m divided by s ના વેગથી નીચે પડે છે, નીચેથી તેના પર ફુંક મારતાં તેનો વેગ  open parentheses 0.20 space i with hat on top space plus space 0.15 space j with hat on top close parentheses space m divided by s થાય છે, તો તેના વેગમાનમાં થતો ફેરફાર.......... kg m/s  હશે.

  • 2 cross times 10 to the power of negative 2 end exponent space i with hat on top space plus space 2 cross times 10 to the power of negative 2 end exponent space j with hat on top
  • 2 cross times 10 to the power of negative 5 end exponent space i with hat on top space plus space 2 cross times 10 to the power of negative 5 end exponent space j with hat on top
  • 2 cross times 10 to the power of negative 2 end exponent space i with hat on top space plus space 1 cross times 10 to the power of negative 2 end exponent space j with hat on top
  • 2 cross times 10 to the power of negative 2 end exponent space i with hat on top space minus space 2 cross times 10 to the power of negative 2 space end exponent space j with hat on top

B.

2 cross times 10 to the power of negative 5 end exponent space i with hat on top space plus space 2 cross times 10 to the power of negative 5 end exponent space j with hat on top

Tips: -

પ્રારંભિક વેગમાન stack P subscript 1 with rightwards arrow on top

 equals space m stack v subscript 1 with rightwards arrow on top
equals space 0.1 space cross times space 10 to the power of negative 3 space end exponent open parentheses negative 0.05 space j with hat on top close parentheses
equals space minus 0.5 space cross times space 10 to the power of negative 5 end exponent space j with hat on top
     
અંતિમ વેગમાન  stack P subscript 2 with rightwards arrow on top

equals space 0.1 space cross times 10 to the power of negative 3 space end exponent open parentheses 0.2 space i with hat on top space plus space 0.15 space j with hat on top close parentheses
equals space 2 space cross times space 10 to the power of negative 5 space end exponent i with hat on top space plus space 1.5 space cross times space 10 to the power of negative 5 end exponent space j with hat on top


વેગમાનમાં ફેરફાર triangle space P with rightwards arrow on top

equals space stack P subscript 2 with rightwards arrow on top space minus space stack P subscript 1 with rightwards arrow on top
equals open parentheses 2 space cross times space 10 to the power of negative 5 space end exponent i with hat on top space plus space 1.5 space cross times space 10 to the power of negative 5 end exponent space j with hat on top close parentheses space minus space open parentheses negative 0.5 space cross times space 10 to the power of negative 5 end exponent space j with hat on top close parentheses
equals space 2 space cross times space 10 to the power of negative 5 end exponent space i with hat on top space plus space 2 space cross times space 10 to the power of negative 5 end exponent space j with hat on top space k g space m divided by s


2 kg નું એક પક્ષી open parentheses 2 space i with hat on top space plus space 4 space j with hat on top close parentheses m/s ના અચળ વેગથી તથા 3 kg નું બીજું પક્ષી  open parentheses 2 space i with hat on top space plus space 6 space j with hat on top close parentheses m/s થી ઊડતાં હોય, તો બંને પક્ષી વડે બનતાં તંત્રના દ્રવ્યમાન-કેન્દ્રનો વેગ...... m/s હોય ?

  • open parentheses 2 space i with hat on top space plus space 5.2 space j with hat on top close parentheses
  • 2 space i with hat on top space plus space 2 space j with hat on top
  • 2 space i with hat on top space minus space 2 space j with hat on top
  • 10 space i with hat on top space plus space 10 space j with hat on top

B.

2 space i with hat on top space plus space 2 space j with hat on top

Tips: -

stack v subscript c m end subscript with rightwards arrow on top equals space fraction numerator m subscript 1 space stack v subscript 1 with rightwards arrow on top space plus space m subscript 2 space stack v subscript 2 with rightwards arrow on top over denominator m subscript 1 space plus space m subscript 2 end fraction space space space

       equals space fraction numerator open parentheses 2 close parentheses space open parentheses 2 space i with hat on top space minus space 4 space j with hat on top close parentheses space plus space open parentheses 3 close parentheses space open parentheses 2 space i with hat on top space plus space 6 space j with hat on top close parentheses over denominator 2 space plus space 3 end fraction

equals space fraction numerator 10 space i with hat on top space plus space 10 space j with hat on top over denominator 5 end fraction

equals space 2 space i with hat on top space plus space 2 space j with hat on top space space m divided by s

જો 't' સમયે કોઈ પથ્થરનું વેગમાન open square brackets open parentheses 0.5 space k g space m divided by s squared close parentheses t to the power of 2 space end exponent plus space open parentheses 3.0 space k g space m divided by s close parentheses close square brackets i with hat on top space plus space open square brackets 1.5 space k g space m divided by s squared close square brackets t space j with hat on top space space right square bracket હોય, તો તેના પર લાગતું બળ કેટલું હોય ?

  • open parentheses t space i with hat on top space plus space 1.5 space j with hat on top close parentheses space N
  • open parentheses 0.5 t i with hat on top space plus space 1.5 j with hat on top close parentheses space N
  • open square brackets open parentheses 0.5 t space plus space 3 close parentheses i with hat on top space plus space 1.5 space j with hat on top close square brackets space N
  • open parentheses 0.5 space i with hat on top space plus space 1.5 j with hat on top close parentheses space N

A.

open parentheses t space i with hat on top space plus space 1.5 space j with hat on top close parentheses space N

Tips: -

P with rightwards arrow on top space equals space open parentheses 0.5 space t to the power of 2 space end exponent plus space 3 close parentheses space space i with hat on top space plus space open parentheses 1.5 t close parentheses space j with hat on top

therefore space P with rightwards arrow on top space equals space stack fraction numerator d p over denominator d t end fraction with rightwards arrow on top space equals space open parentheses 0.5 space cross times 2 t close parentheses space i with hat on top space plus space 1.5 space j with hat on top
                   equals space open parentheses t space stack i space with hat on top space plus space 1.5 space j with hat on top close parentheses space N

ધારો કે તમારું દ્રવ્યમાન 50 kg છે. તમારે કેટલી ઝડપથી દોડવું પડે કે જેથી તમારું રેખીય વેગમાન 20 km/h ની ઝડપથી સીધા રસ્તા પર ગતિ કરતા 100 kg સાઇકલ સવાર જેટલું થાય ? 

  • 40m/s

  • 11.11 m/s

  • 20 km/h

  • 10 km/h


B.

11.11 m/s

Tips: -

સાઇકલનો વેગ  =
 nu subscript b space end subscript equals space 20 space k m divided by h
space space space space equals space fraction numerator 20 cross times 1000 over denominator 3600 end fraction space equals space 5.55 space m space s to the power of negative 1 end exponent

હવે, વ્યક્તિનું વેગમાન = સાઇકલનું વેગમાન

therefore space m v space equals space m subscript b space end subscript times space v subscript b

therefore space left parenthesis 50 right parenthesis space v space equals space left parenthesis 100 right parenthesis space left parenthesis 5.55 right parenthesis

therefore space v space equals space fraction numerator 100 cross times 5.55 over denominator 50 end fraction space equals space 11.1 space m space s to the power of negative 1 end exponent


2400 kgની એક બસ સીધા રસ્તા પર્ 60 km/hની ઝડપથી જાય છે. બસની પાછળ 1600 kgની એક કાર 80 km/h ની ઝડપથી આવી રહી છે. બંને વાહનોનુ દ્રવ્યમાન-કેન્દ્ર કેટલી ઝડપથી ગતિ કરતું હશે ?

  • 70 km/h

  • 75 km/h

  • 72 km/h

  • 68 km/h


D.

68 km/h

Tips: -

દ્રવ્યમાન-કેન્દ્રનો વેગ,

v subscript c m space end subscript equals space fraction numerator m subscript 1 v subscript 1 space end subscript plus space m subscript 2 v subscript 2 over denominator m subscript 1 space end subscript plus space m subscript 2 end fraction

equals fraction numerator left parenthesis 2400 cross times 60 right parenthesis space plus space left parenthesis 1600 cross times 80 right parenthesis over denominator 2400 space plus space 1600 end fraction space equals space 68 space k m divided by h