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The cause of quantization of electric charge is
The quantization of electric charge occurs because charge is transferred in discrete units of the elementary charge e, which is the charge of an electron or proton. This means charge always exists in integral multiples of e, never in fractions. Answer: (c) transfer of integral number of electrons. FRead more
The quantization of electric charge occurs because charge is transferred in discrete units of the elementary charge e, which is the charge of an electron or proton. This means charge always exists in integral multiples of e, never in fractions.
Answer: (c) transfer of integral number of electrons.
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On rationalizing the denominator of 1√2, we get
We are given the expression: 1/√2 To rationalize the denominator, we multiply both the numerator and denominator by √2 to eliminate the square root in the denominator Step 1: Multiply by √2/√2 1/√2 × √2/√2 = √2/2 Conclusion: The correct answer is √2/2 (option d). This question related to Chapter 1Read more
We are given the expression: 1/√2
To rationalize the denominator, we multiply both the numerator and denominator by √2 to eliminate the square root in the denominator
Step 1: Multiply by √2/√2
1/√2 × √2/√2 = √2/2
Conclusion: The correct answer is √2/2 (option d).
This question related to Chapter 1 Mathematics Class 9th NCERT. From the Chapter 1 Number System. Give answer according to your understanding.
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Decimal expansion of a rational number is terminating if in its denominator there is:
A rational number p/q has a terminating decimal expansion if and only if the prime factorization of its denominator q (after simplify the fraction) contains only the prime factorization of its denominator q (after simplifying the fraction) contains only the prime factor 2 and/or 5. (a). 2 or 5 IfRead more
A rational number p/q has a terminating decimal expansion if and only if the prime factorization of its denominator q (after simplify the fraction) contains only the prime factorization of its denominator q (after simplifying the fraction) contains only the prime factor 2 and/or 5.
(a). 2 or 5
If the denominator has only 2, and 5, or both(e.g., 10, 20, 25, etc.), the decimal expansion terminates.
(b). 3 or 5
The presence of 3 in the denominator(e.g., 1/3 = 0.3333…) causes a non-terminating repeating decimal.
(c). 9 or 11
9 = 3³ and 11 are not factors that produce a terminating decimal.
(d) 3 or 7
Both 3 and 7 cause non-terminating repeating decimals.
This question related to Chapter 1 Mathematics Class 9th NCERT. From the Chapter 1 Number System. Give answer according to your understanding.
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Which of the following is true?
Let's analyze each statement carefully: Option (a): Every whole number is a natural number Natural numbers (N): {1,2, 3, 4,...} Whole numbers (W): {0,1,2,3,4,...} Since whole number include 0, but natural number do not, this statement is false. Option (b): Every integer is a rational number IntegersRead more
Let’s analyze each statement carefully:
Option (a): Every whole number is a natural number
Natural numbers (N): {1,2, 3, 4,…}
Whole numbers (W): {0,1,2,3,4,…}
Since whole number include 0, but natural number do not, this statement is false.
Option (b): Every integer is a rational number
Integers (Z): {…, -3, -2, -1, 0, 1,2,3,…)
Rotational number (Q): Numbers that can be expressed as p/q, where p and q are integers, and q ≠ 0.
Every integer x can be written as x/1, which is in the form of a rotational number.
Thus this statement is true.
Option (c): Every rational number is an integer.
Example 1/2 is a rational number but not an integer.
So, this statement is false.
Option (d): Every integer is a whole number
Whole number do not include negative numbers, but integers do.
Example: -1 is an integer but not a whole number.
Thus, this statement is false.
Final Answer: The correct answer is (b) Every integer is a rotational number. This question related to Chapter 1 Mathematics Class 9th NCERT. From the Chapter 1 Number System. Give answer according to your understanding.
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If m and n are two natural numbers and mⁿ = 32, then nᵐⁿ is
We are given that m and n are two natural numbers, and mⁿ = 32 We need to find the value of nᵐⁿ. Step 1: Find Possible Value of m and n Since 32 and 2⁵, we look for natural numbers m and n such that: mⁿ = 32 Possible values are: m = 2, n = 5( since 2⁵ = 32) Step 2: Compute nᵐⁿ We substitute n = 5Read more
We are given that m and n are two natural numbers, and mⁿ = 32
We need to find the value of nᵐⁿ.
Step 1: Find Possible Value of m and n
Since 32 and 2⁵, we look for natural numbers m and n such that: mⁿ = 32
Possible values are:
m = 2, n = 5( since 2⁵ = 32)
Step 2: Compute nᵐⁿ
We substitute n = 5 and mⁿ = 32:
nᵐⁿ = 5³²
Step 3: Identify the correct option
None of the options match 5³² directly.
Please verify if there was a misinterpretation in the problem statement or options. If you meant nᵐ*ⁿ instead of nᵐⁿ , let’s check:
nᵐ*ⁿ = 5²×⁵ = 5¹⁰
This matches option (c) 5¹⁰.
This question related to Chapter 1 Mathematics Class 9th NCERT. From the Chapter 1 Number System. Give answer according to your understanding.
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