B. कृषि भूमि के निजीकरण और मजदूरों के शहरों की ओर प्रवास/Privatization of farmland and movement of labourers to towns
Step 1
Concept
Enclosure affected small farmers and pushed labourers toward towns. In exams, link it with industrial labour supply.
Step 2
Why this answer is correct
The correct answer is B. कृषि भूमि के निजीकरण और मजदूरों के शहरों की ओर प्रवास / Privatization of farmland and movement of labourers to towns. Enclosure affected small farmers and pushed labourers toward towns. In exams, link it with industrial labour supply.
Step 3
Exam Tip
एनक्लोजर ने छोटे किसानों को प्रभावित कर श्रमिकों को शहरों की ओर धकेला। परीक्षा में इसे औद्योगिक श्रम आपूर्ति से जोड़ें।
B. इसने कई ग्रामीण लोगों को मजदूरी काम की ओर धकेला/It pushed many rural people toward wage work
Step 1
Concept
Enclosure pushed part of the rural population toward cities and wage labor. For exams see the link between agrarian change and industry.
Step 2
Why this answer is correct
The correct answer is B. इसने कई ग्रामीण लोगों को मजदूरी काम की ओर धकेला / It pushed many rural people toward wage work. Enclosure pushed part of the rural population toward cities and wage labor. For exams see the link between agrarian change and industry.
Step 3
Exam Tip
एन्क्लोजर से ग्रामीण आबादी का हिस्सा शहरों और मजदूरी श्रम की ओर गया। परीक्षा में कृषि परिवर्तन और उद्योग संबंध देखें।
A. कई ग्रामीण लोग मजदूरी और शहरों की ओर धकेले गए/Many rural people were pushed toward wage labor and towns
Step 1
Concept
Enclosure weakened many small peasants and increased wage labor. For exams connect it with urbanization.
Step 2
Why this answer is correct
The correct answer is A. कई ग्रामीण लोग मजदूरी और शहरों की ओर धकेले गए / Many rural people were pushed toward wage labor and towns. Enclosure weakened many small peasants and increased wage labor. For exams connect it with urbanization.
Step 3
Exam Tip
एनक्लोजर से छोटे किसानों की स्थिति कमजोर हुई और मजदूरी श्रम बढ़ा। परीक्षा में इसे शहरीकरण से जोड़ें।
Lotus Mahal is considered associated with the Zenana enclosure. In exams understand it as a Vijayanagara building with Indo-Islamic influence.
Step 2
Why this answer is correct
The correct answer is A. जनाना परिसर / Zenana enclosure. Lotus Mahal is considered associated with the Zenana enclosure. In exams understand it as a Vijayanagara building with Indo-Islamic influence.
Step 3
Exam Tip
लोटस महल जनाना परिसर से जुड़ा माना जाता है। परीक्षा में इसे इंडो इस्लामी प्रभाव वाली विजयनगर इमारत समझें।
This is the sum of (16) terms with \(t_5=38\) and \(t_{20}=128\). Exam tip: treat the required middle part as a smaller AP.
Step 2
Why this answer is correct
The correct answer is D. (1328). This is the sum of (16) terms with \(t_5=38\) and \(t_{20}=128\). Exam tip: treat the required middle part as a smaller AP.
Step 3
Exam Tip
यह योग (16) पदों का है जिसमें \(t_5=38\) और \(t_{20}=128\) हैं। परीक्षा में बीच के पदों का योग छोटे भाग के रूप में निकालें।
The required sum is \(S_{28}-S_{14}=1806\). The sum of consecutive terms is quickly found by subtracting partial sums.
Step 2
Why this answer is correct
The correct answer is B. (1806). The required sum is \(S_{28}-S_{14}=1806\). The sum of consecutive terms is quickly found by subtracting partial sums.
Step 3
Exam Tip
आवश्यक योग \(S_{28}-S_{14}=1806\) है। लगातार पदों का योग आंशिक योगों के अंतर से तुरंत मिलता है।
The required sum is \(S_{24}-S_{12}=1104\). The sum of consecutive terms is quickly found by subtracting partial sums.
Step 2
Why this answer is correct
The correct answer is B. (1104). The required sum is \(S_{24}-S_{12}=1104\). The sum of consecutive terms is quickly found by subtracting partial sums.
Step 3
Exam Tip
आवश्यक योग \(S_{24}-S_{12}=1104\) है। लगातार पदों का योग आंशिक योगों के अंतर से तुरंत मिलता है।
The required sum is \(S_{30}-S_{12}=1566\). To find a middle block sum, subtract the previous partial sum from the larger sum.
Step 2
Why this answer is correct
The correct answer is B. (1566). The required sum is \(S_{30}-S_{12}=1566\). To find a middle block sum, subtract the previous partial sum from the larger sum.
Step 3
Exam Tip
मांगा गया योग \(S_{30}-S_{12}=1566\) है। बीच के पदों का योग निकालने के लिए बड़े योग से पहले वाला योग घटाएँ।
After cancellation, the denominator becomes \(2^4\cdot 5^2\cdot 17\). Since (17) remains, the decimal is non-terminating recurring.
Step 2
Why this answer is correct
The correct answer is B. असांत आवर्ती / Non-terminating recurring. After cancellation, the denominator becomes \(2^4\cdot 5^2\cdot 17\). Since (17) remains, the decimal is non-terminating recurring.
Step 3
Exam Tip
कटौती के बाद हर \(2^4\cdot 5^2\cdot 17\) बचेगा। (17) बचने से दशमलव असांत आवर्ती होगा।
Since \(320=2^6\cdot 5\), the reduced denominator is \(2\cdot 5^2\cdot 11\). Since (11) remains, the decimal is non-terminating recurring.
Step 2
Why this answer is correct
The correct answer is B. असांत आवर्ती / Non-terminating recurring. Since \(320=2^6\cdot 5\), the reduced denominator is \(2\cdot 5^2\cdot 11\). Since (11) remains, the decimal is non-terminating recurring.
Step 3
Exam Tip
\(320=2^6\cdot 5\) कटने पर हर \(2\cdot 5^2\cdot 11\) बचेगा। (11) बचने से दशमलव असांत आवर्ती होगा।
After cancelling \(22=2\cdot 11\), the denominator becomes \(2\cdot 5^4\cdot 11\). Since (11) remains, the decimal is non-terminating recurring.
Step 2
Why this answer is correct
The correct answer is B. असांत आवर्ती / Non-terminating recurring. After cancelling \(22=2\cdot 11\), the denominator becomes \(2\cdot 5^4\cdot 11\). Since (11) remains, the decimal is non-terminating recurring.
Step 3
Exam Tip
\(22=2\cdot 11\) कटने पर हर \(2\cdot 5^4\cdot 11\) बचेगा। (11) बचने से दशमलव असांत आवर्ती होगा।
Since \(245=5\cdot 7^2\), the reduced denominator is \(2^2\cdot 5\cdot 7\). Since (7) remains, the decimal is non-terminating recurring.
Step 2
Why this answer is correct
The correct answer is C. असांत आवर्ती / Non-terminating recurring. Since \(245=5\cdot 7^2\), the reduced denominator is \(2^2\cdot 5\cdot 7\). Since (7) remains, the decimal is non-terminating recurring.
Step 3
Exam Tip
\(245=5\cdot 7^2\) कटने पर हर \(2^2\cdot 5\cdot 7\) बचता है। (7) बचने से दशमलव असांत आवर्ती होगा।
After cancelling \(55=5\cdot 11\), the denominator becomes \(2^2\cdot 5^2\cdot 11\). Since (11) remains, the decimal is non-terminating recurring.
Step 2
Why this answer is correct
The correct answer is B. असांत आवर्ती / Non-terminating recurring. After cancelling \(55=5\cdot 11\), the denominator becomes \(2^2\cdot 5^2\cdot 11\). Since (11) remains, the decimal is non-terminating recurring.
Step 3
Exam Tip
\(55=5\cdot 11\) कटने पर हर \(2^2\cdot 5^2\cdot 11\) बचेगा। (11) बचने से दशमलव असांत आवर्ती होगा।
After cancellation, the denominator is \(2^3\cdot 3\cdot 5^4\cdot 11\), which contains (3) and (11). If primes other than (2) and (5) remain in the reduced denominator, the decimal is non-terminating recurring.
Step 2
Why this answer is correct
The correct answer is B. असांत आवर्ती / Non-terminating recurring. After cancellation, the denominator is \(2^3\cdot 3\cdot 5^4\cdot 11\), which contains (3) and (11). If primes other than (2) and (5) remain in the reduced denominator, the decimal is non-terminating recurring.
Step 3
Exam Tip
कटौती के बाद हर \(2^3\cdot 3\cdot 5^4\cdot 11\) बचता है, जिसमें (3) और (11) हैं। सरलतम हर में (2) और (5) के अलावा गुणनखंड बचें तो दशमलव असांत आवर्ती होता है।
After cancellation, the denominator becomes \(2^3\cdot 5^3\cdot 13\). Since (13) remains, the decimal is non-terminating recurring.
Step 2
Why this answer is correct
The correct answer is B. असांत आवर्ती / Non-terminating recurring. After cancellation, the denominator becomes \(2^3\cdot 5^3\cdot 13\). Since (13) remains, the decimal is non-terminating recurring.
Step 3
Exam Tip
कटौती के बाद हर \(2^3\cdot 5^3\cdot 13\) बचेगा। (13) बचने से दशमलव असांत आवर्ती होगा।
Since \(200=2^3\cdot 5^2\), the reduced denominator is \(5\cdot 7\). Since (7) remains, the decimal is non-terminating recurring.
Step 2
Why this answer is correct
The correct answer is B. असांत आवर्ती / Non-terminating recurring. Since \(200=2^3\cdot 5^2\), the reduced denominator is \(5\cdot 7\). Since (7) remains, the decimal is non-terminating recurring.
Step 3
Exam Tip
\(200=2^3\cdot 5^2\) कटने पर हर \(5\cdot 7\) बचेगा। (7) बचने से दशमलव असांत आवर्ती होगा।
After cancelling \(14=2\cdot 7\), the denominator becomes \(2\cdot 5^3\cdot 7\). Since (7) remains, the decimal is non-terminating recurring.
Step 2
Why this answer is correct
The correct answer is B. असांत आवर्ती / Non-terminating recurring. After cancelling \(14=2\cdot 7\), the denominator becomes \(2\cdot 5^3\cdot 7\). Since (7) remains, the decimal is non-terminating recurring.
Step 3
Exam Tip
\(14=2\cdot 7\) कटने पर हर \(2\cdot 5^3\cdot 7\) बचेगा। (7) बचने से दशमलव असांत आवर्ती होगा।
Since \(175=5^2\cdot 7\), the reduced denominator is \(2^2\cdot 5\cdot 7\). Since (7) remains, the decimal is non-terminating recurring.
Step 2
Why this answer is correct
The correct answer is C. असांत आवर्ती / Non-terminating recurring. Since \(175=5^2\cdot 7\), the reduced denominator is \(2^2\cdot 5\cdot 7\). Since (7) remains, the decimal is non-terminating recurring.
Step 3
Exam Tip
\(175=5^2\cdot 7\) कटने पर हर \(2^2\cdot 5\cdot 7\) बचता है। (7) बचने से दशमलव असांत आवर्ती होगा।
After cancellation, the denominator becomes \(2^3\cdot 5^2\cdot 7\). Since (7) remains, the decimal is non-terminating recurring.
Step 2
Why this answer is correct
The correct answer is B. असांत आवर्ती / Non-terminating recurring. After cancellation, the denominator becomes \(2^3\cdot 5^2\cdot 7\). Since (7) remains, the decimal is non-terminating recurring.
Step 3
Exam Tip
कटौती के बाद हर \(2^3\cdot 5^2\cdot 7\) बचेगा। (7) बचने से दशमलव असांत आवर्ती होगा।
After cancelling \(14=2\cdot 7\), the denominator becomes \(5^2\cdot 7\). Since (7) remains, the decimal is non-terminating recurring.
Step 2
Why this answer is correct
The correct answer is B. असांत आवर्ती / Non-terminating recurring. After cancelling \(14=2\cdot 7\), the denominator becomes \(5^2\cdot 7\). Since (7) remains, the decimal is non-terminating recurring.
Step 3
Exam Tip
\(14=2\cdot 7\) कटने पर हर \(5^2\cdot 7\) बचेगा। (7) बचने से दशमलव असांत आवर्ती होगा।
Even after \(125=5^3\) cancels, (11) remains in the denominator. If a reduced denominator has a prime other than (2) and (5), the decimal is non-terminating recurring.
Step 2
Why this answer is correct
The correct answer is B. असांत आवर्ती / Non-terminating recurring. Even after \(125=5^3\) cancels, (11) remains in the denominator. If a reduced denominator has a prime other than (2) and (5), the decimal is non-terminating recurring.
Step 3
Exam Tip
\(125=5^3\) कटने पर भी हर में (11) बचता है। सरलतम हर में (2) और (5) के अलावा कोई अभाज्य रहे तो दशमलव असांत आवर्ती होता है।
After cancellation, the denominator becomes \(5\cdot 7\). Since (7) remains, the decimal is non-terminating recurring.
Step 3
Exam Tip
Check whether the whole power cancels or only part of it cancels. चरण 1: \(98=2\cdot 7^2\) है। चरण 2: कटौती के बाद हर \(5\cdot 7\) बचेगा। (7) बचने से दशमलव असांत आवर्ती होगा। चरण 3: घात पूरी कटे या नहीं, यह ध्यान से देखें।
The reduced denominator becomes \(2^3\cdot 3\cdot 5^2\). Since (3) remains, the decimal is non-terminating recurring.
Step 3
Exam Tip
A prime factor may cancel only partially. चरण 1: अंश से \(2^4\cdot 3\) कटेगा। चरण 2: सरलतम हर \(2^3\cdot 3\cdot 5^2\) बचेगा। इसमें (3) बचा है, इसलिए दशमलव असांत आवर्ती होगा। चरण 3: एक ही अभाज्य गुणनखंड आंशिक रूप से कट सकता है।
The numerator (13) cancels only one factor (13) from \(13^2\).
Step 2
Why this answer is correct
The reduced denominator is \(2^2\cdot 5^2\cdot 13\). Since (13) remains, the decimal is non-terminating recurring.
Step 3
Exam Tip
Understand the difference between complete and partial cancellation. चरण 1: अंश का (13) हर के \(13^2\) में से केवल एक (13) काटेगा। चरण 2: सरलतम हर \(2^2\cdot 5^2\cdot 13\) बचेगा। (13) बचने से दशमलव असांत आवर्ती होगा। चरण 3: पूरी और आंशिक कटौती में फर्क समझें।
After cancellation, the denominator becomes \(2\cdot 5\cdot 11\). Since (11) remains, the decimal is non-terminating recurring.
Step 3
Exam Tip
After partial cancellation, always check the remaining factors. चरण 1: \(55=5\cdot 11\) है। चरण 2: कटौती के बाद हर \(2\cdot 5\cdot 11\) बचेगा। (11) बचने के कारण दशमलव असांत आवर्ती होगा। चरण 3: आंशिक कटौती के बाद बचे हुए गुणनखंडों को जरूर जाँचें।
The factor (5) and one (7) cancel, but one (7) remains. The reduced denominator is \(2^2\cdot 7\). So the decimal is non-terminating recurring.
Step 3
Exam Tip
After partial cancellation, check what factor remains. चरण 1: \(35=5\cdot 7\) है। चरण 2: हर से (5) और एक (7) कटेगा, पर एक (7) बच जाएगा। सरलतम हर \(2^2\cdot 7\) है। इसलिए दशमलव असांत आवर्ती होगा। चरण 3: आंशिक कटौती के बाद बचे गुणनखंड को जरूर देखें।
A one-unit rise in pH decreases acidity ten times.
Step 2
Why this answer is correct
pH has risen two units from 3 to 5.
Step 3
Exam Tip
Therefore acidity decreased about one hundred times. चरण 1: पीएच में एक इकाई वृद्धि अम्लीयता को दस गुना घटाती है। चरण 2: पीएच तीन से पाँच तक दो इकाई बढ़ा है। चरण 3: इसलिए अम्लीयता लगभग सौ गुना घट गई।
A. अम्लीयता कम हुई पर विलयन अभी भी अम्लीय है/Acidity decreased but the solution is still acidic
Step 1
Concept
pH 2 and pH 3 are both below 7.
Step 2
Why this answer is correct
Increase in pH means fewer hydrogen ions.
Step 3
Exam Tip
Therefore acidity decreased but the solution remained acidic. चरण 1: पीएच दो और तीन दोनों सात से कम हैं। चरण 2: पीएच बढ़ने का अर्थ हाइड्रोजन आयन की कमी है। चरण 3: इसलिए अम्लीयता कम हुई लेकिन विलयन अम्लीय ही रहा।
A. अम्लीयता घटी पर विलयन अभी भी अम्लीय है/Acidity decreased but the solution is still acidic
Step 1
Concept
pH 3 and pH 4 are both below 7, so the solution is acidic.
Step 2
Why this answer is correct
Increase in pH means fewer hydrogen ions.
Step 3
Exam Tip
Therefore acidity decreased but the solution is still acidic. चरण 1: पीएच तीन और चार दोनों सात से कम हैं इसलिए विलयन अम्लीय है। चरण 2: पीएच बढ़ने से हाइड्रोजन आयन घटते हैं। चरण 3: इसलिए अम्लीयता कम हुई पर विलयन अभी भी अम्लीय है।
A. सभी हाइड्रोजन आयन उदासीन नहीं हुए/Not all hydrogen ions have been neutralised
Step 1
Concept
A base reduces hydrogen ions of an acid.
Step 2
Why this answer is correct
A small amount of base neutralises only some hydrogen ions.
Step 3
Exam Tip
Therefore pH increases but the solution is not fully neutralised. चरण 1: क्षार अम्ल के हाइड्रोजन आयनों को घटाता है। चरण 2: थोड़े क्षार से केवल कुछ हाइड्रोजन आयन ही उदासीन होते हैं। चरण 3: इसलिए पीएच बढ़ता है लेकिन विलयन पूरी तरह उदासीन नहीं होता।
