Fill in the blanks by suitable conversion of units (a) 1 kg m2 s-2 = �. g cm2 s-2 (b) 1 m =���� ly (c) 3.0 m s-2 = �. km h-2 (d) G = 6.67 x 10-11 N m2 (kg)-2 = �. (cm)3 s-2 g-1

Asked by Last Modified  

Follow 21
Answer

Please enter your answer

Tutor

Q(a)10^7 Q(b)1.06*10^-16 Q(c)10.8 Q(d)6.67*10^-8
Comments

JEE Main Maths, Physics With 12 years exp.

(a) 10^7 (b)1.06×10^-16 (C)10.8 (d)6.67×10^-8
Comments

Tutor

(a) 1 kg = 103 g1 m2 = 104 cm21 kg m2 s–2 = 1 kg × 1 m2 × 1 s–2=103 g × 104 cm2 × 1 s–2 = 107 g cm2 s–21 kg m2s–2= 107 g cm2 s–2 (b) Distance = Speed × TimeSpeed of light = 3 × 108 m/sTime = 1 year = 365 days = 365 × 24 hours...
read more
(a) 1 kg = 103g1 m2= 104cm21 kg m2s–2= 1 kg × 1 m2× 1 s–2=103g × 104cm2× 1 s–2= 107g cm2s–21 kg m2s–2=107g cm2s–2 (b)Distance = Speed × TimeSpeed of light = 3 × 108m/sTime = 1 year = 365 days = 365 × 24 hours = 365 × 24 × 60 × 60 secPutting these values in above formula we get1 light year distance = (3 × 108m/s) × (365 × 24 × 60 × 60 s) = 9.46 × 1015m9.46 × 1015m = 1 lySo that 1 m = 1/ 9.46 × 1015ly =1.06 × 10–16ly (c) 1 hour = 3600 sec so that 1 sec = 1/3600 hour1 km = 1000 m so that 1 m = 1/1000 km3.0 m s–2= 3.0 (1/1000 km)( 1/3600 hour)-2= 3.0 × 10–3km × ((1/3600)-2h–2)= 3.0 × 10–3km × (3600)2h–2= 3.88 × 104km h–23.0 m s–2=3.88 × 104km h–2(d)Given,G= 6.67 × 10–11N m2kg–2We know that1 N = 1 kg m s–21 kg = 103g1 m = 100 cm = 102cmPutting above values, we get6.67 × 10–11N m2kg–2= 6.67 × 10–11× (1 kg m s–2) (1 m2) (1Kg–2)Solve and cancel out the units, we get⇒ 6.67 × 10–11× (1 kg–1× 1 m3× 1 s–2)Putting above values to convert kg to g and m to cm⇒ 6.67 × 10–11× (103g)-1× (102cm)3× (1 s–2)⇒ 6.67 × 10–11× 10-3g-1× 106cm3× (1 s–2)⇒ 6.67 × 10–8cm3s–2g–1G= 6.67 × 10–11N m2kg–2 =6.67 × 10–8cm3s–2g–1 read less
Comments

Question (a) 10^7 Question (b)1.06×10^-16 Question (c)10.8 Question (d)6.67×10^-8
Comments

(a) 10^7 (b)1.06×10^-16 (C)10.8 (d)6.67×10^-8
Comments

(a) 10^7 (b)1.06×10^-16 (C)10.8 (d)6.67×10^-8
Comments

B. TECH.

Q(a)10^7 Q(b)1.06*10^-16 Q(c)10.8 Q(d)6.67*10^-8
Comments

Excellent track record tutor

Q(a)10^7 Q(b)1.06*10^-16 Q(c)10.8 Q(d)6.67*10^-8
Comments

Tutor

(a)10^7 (b)1.06*10^-16 (c)10.8 (d)6.67*10^-8
Comments

Tutor with teaching experience of 15+ years..

(a) 10^7 (b)1.06×10^-16 (C)10.8 (d)6.67×10^-8
Comments

View 22 more Answers

Related Questions

Which books are best in a Physical Education for CBSENET exam?
Saraswati is good book. Vishvas publication is also good. Dear student please please please study physical education from beginning because it is going to be very tough marking from 2017 onwards many students got failed.
Pandu
The emf of a source is equal to the potential difference across the terminals of the source when either its internal resistance is __________. (Zero, Infinite, Zero or infinite, None of these)
When the internal resistance is zero. If otherwise, the voltage across the terminals of the source suffers a loss across the internal resistance of the emf source.
Narendra Kumar
1 0
7
What is the life span of an elephant?
Asian Elephant = 48 years African bush Elephant = 60-70 years. African Forest Elephant = 60-70 years.
Navneet
0 0
5
Find out the probability on obtaining an even prime number on each die, when a pair of dice is rolled?
There is only one case (2,2) so probability = 1/36
Kumar Upendra Akshay
1 0
6
I am teacher of accountancy . I want to start teach student online . How can I start ?????
you can start by joining Online coaching institutes as a faculty, there is a great scope after that
geeta educations

Now ask question in any of the 1000+ Categories, and get Answers from Tutors and Trainers on UrbanPro.com

Ask a Question

Related Lessons

Did you know?
PERIODIC TABLE- Periodic Laws- Genesis of Periodic Classification After the proposal of John Dalton's Atomic Theory, the scientists took Atomic Weight as the important property of element and tried...

Lesson 1: Multiplication Trick: Multiply Quickly Without Any Error
Vedic Maths: Multiplication tips and tricks Introduction : Founder of Vedic maths is Swami Bharti Krishna Tirath jee Maharaj who was a great saint. With the help of these tri Nikhilam Sutra(Base...


Easy Way to Exam preparation
List out the chapters that you are studied and easy for you. List out the chapter that covers max marks and easy. List out sub topics names in each chapter and prepare each concept by understanding...

Chemistry- Lewis Concept Of Acids And Bases.
Explain the Lewis concept of acids and bases.A Lewis acid is, therefore, any substance, such as the H+ ion, that can accept a pair of nonbonding electrons. In other words, a Lewis acid is an electron-pair...

Recommended Articles

Sandhya is a proactive educationalist. She conducts classes for CBSE, PUC, ICSE, I.B. and IGCSE. Having a 6-year experience in teaching, she connects with her students and provides tutoring as per their understanding. She mentors her students personally and strives them to achieve their goals with ease. Being an enthusiastic...

Read full article >

Raghunandan is a passionate teacher with a decade of teaching experience. Being a skilled trainer with extensive knowledge, he provides high-quality BTech, Class 10 and Class 12 tuition classes. His methods of teaching with real-time examples makes difficult topics simple to understand. He explains every concept in-detail...

Read full article >

Swati is a renowned Hindi tutor with 7 years of experience in teaching. She conducts classes for various students ranging from class 6- class 12 and also BA students. Having pursued her education at Madras University where she did her Masters in Hindi, Swati knows her way around students. She believes that each student...

Read full article >

Radhe Shyam is a highly skilled accounts and finance trainer with 8 years of experience in teaching. Accounting is challenging for many students and that’s where Radhe Shyam’s expertise comes into play. He helps his students not only in understanding the subject but also advises them on how to overcome the fear of accounts...

Read full article >

Looking for Class 12 Tuition ?

Learn from the Best Tutors on UrbanPro

Are you a Tutor or Training Institute?

Join UrbanPro Today to find students near you